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PHYSICS, M5 2023 VCE 2

Phobos is a small moon in a circular orbit around Mars at an altitude of 6000 km above the surface of Mars.

The gravitational field strength of Mars at its surface is 3.72 N kg\(^{-1}\). The radius of Mars is 3390 km.

  1. Show that the gravitational field strength 6000 km above the surface of Mars is 0.48 N kg\(^{-1}\).   (2 marks)
  1. Calculate the orbital period of Phobos. Give your answer in seconds.  (3 marks)
  1. Phobos is very slowly getting closer to Mars as it orbits.
  2. Will the orbital period of Phobos become shorter, stay the same or become longer as it orbits closer to Mars? Explain your reasoning.  (2 marks)

Show Answers Only

a.    \(0.48\ \text{N kg}^{-1}\)

b.    \(2.77 \times 10^4\ \text{s}\)

c.    The orbital period will become shorter.

Show Worked Solution

a.    \(\text{Find the mass of Mars:}\)

         \(g\) \(=\dfrac{GM}{r^2}\)
  \(M\) \(=\dfrac{gr^2}{G}\)
    \(=\dfrac{3.72 \times (3390 \times 10^3)^2}{6.67 \times 10^{-11}}\)
    \(=6.409 \times 10^{23}\)

 
\(\text{Find the gravitational field strength where}\ \ r= 9390\ \text{km:}\) 

         \(g\) \(=\dfrac{GM}{r^2}\)
    \(=\dfrac{6.67 \times 10^{-11} \times 6.41 \times 10^{23}}{(9\ 390\ 000)^2}\)
    \(=0.48\ \text{N kg}^{-1}\)

 

b.     \(\dfrac{r^3}{T^2}\) \(=\dfrac{GM}{4\pi^2}\)
  \(T\) \(=\sqrt{\dfrac{4\pi^2 r^3}{GM}}\)
    \(=\sqrt{\dfrac{4\pi^2 (9\ 390\ 000)^3}{6.67 \times 10^{-11} \times 6.409 \times 10^{23}}}\)
    \(=27\ 652\ \text{s}\)
    \(=2.77 \times 10^4\ \text{s}\)

♦ Mean mark (b) 53%. 
COMMENT: Rounded calculations, such as using 6.41 × 10^23 were marked as wrong.

c.    \(T=\sqrt{\dfrac{4\pi^2 r^3}{GM}}\)

\(T \propto \sqrt{r^3}\)

→ \(\text{decreasing the orbital radius will also decrease the orbital period.}\)

PHYSICS, M5 2023 VCE 9*

An engineer is designing a banked circular curve of radius 25 m in a new bicycle velodrome.

Diagram A shows the bicycle approaching the banked section, and diagram B shows the front view of a bicycle moving out of the page as it rounds the banked bend.
 

The bicycle is travelling at 11 m s\(^{-1}\) on the banked section. At this speed there are no sideways frictional forces between the wheels and the road surface.

Determine the angle of the banked bend with the road surface, giving your answer to the nearest degree.   (3 marks)

Show Answers Only

\(26^{\circ}\)

Show Worked Solution

\(\tan \theta\) \(=\dfrac{F_c}{mg}\)  
\(F_c\) \(=mg \tan \theta\)  
\(\dfrac{mv^2}{r}\) \(=mg \tan \theta\)  
\(\dfrac{v^2}{r}\) \(=g \tan \theta\)  
\(\tan \theta\) \(=\dfrac{v^2}{gr}\)  
  \(=\dfrac{11^2}{9.8 \times 25}\)  
\(\theta\) \(=\tan^{-1}\Big(\dfrac{11^2}{9.8 \times 25}\Big)\)  
  \(=26^{\circ}\)  

PHYSICS, M6 2023 VCE 5-6 MC

The diagram below shows a stationary circular coil of conducting wire connected to a low-resistance globe in a uniform, constant magnetic field, \(B\).
 

 

Question 5

The magnetic field is switched off.

Which one of the following best describes the globe in the circuit \( \textbf{before} \) the magnetic field is switched off, \( \textbf{during} \) the time the magnetic field is being switched off and \( \textbf{after} \) the magnetic field is switched off?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad\text{Before}\quad\rule[-1ex]{0pt}{0pt}&\quad \text{During} \quad& \quad\text{After}\quad\\
\hline
\rule{0pt}{2.5ex}\text{Off}\rule[-1ex]{0pt}{0pt}&\text{On}& \text{Off}\\
\hline
\rule{0pt}{2.5ex}\text{On}\rule[-1ex]{0pt}{0pt}& \text{On}& \text{Off}\\
\hline
\rule{0pt}{2.5ex}\text{On}\rule[-1ex]{0pt}{0pt}& \text{Off} & \text{Off}\\
\hline
\rule{0pt}{2.5ex}\text{Off}\rule[-1ex]{0pt}{0pt}& \text{On} & \text{On}\\
\hline
\end{array}
\end{align*}

Question 6

The radius of the coil is 5 cm and the magnetic field strength is 0.2 T. The coil has 100 loops. Assume that the magnetic field is perpendicular to the area of the coil.

Which one of the following is closest to the magnitude of the magnetic flux through the coil of wire when the magnetic field is switched on?

  1.  0.0016 Wb
  2.  0.16 Wb
  3.  16 Wb
  4. 1600 Wb
Show Answers Only

\(\text{Question 5:}\ A\)

\(\text{Question 6:}\ A\)

Show Worked Solution

\(\text{Question 5}\)

→ By Faraday’s Law of Induction, an EMF and current will be induced in the coil when the coil experiences a change of flux causing the light to turn on. 

→ This occurs when the strength of the magnetic field is changing and so the light will be on during the time the magnetic field is being reduced to 0.

→ Before and after this, there is no change in flux (i.e. no EMF or current produced) so the light will be off.

\(\Rightarrow A\)


\(\text{Question 6}\)

\(\phi=BA=B \times \pi \times r^2 = 0.2 \times \pi \times (0.05)^2= 0.0016\ \text{Wb}\)

  
\(\Rightarrow A\)

PHYSICS, M6 2023 VCE 1 MC

One type of loudspeaker consists of a current-carrying coil within a radial magnetic field, as shown in the diagram below. \(X\) and \(Y\) are magnetic poles, and the direction of the current, \(I\), in the coil is clockwise as shown.
 


 

The force, \(F\), acting on the current-carrying coil is directed into the page.

Which one of the following statements correctly identifies the magnetic polarities of \(X\) and \(Y\)?

  1. \(X\) is a north pole and \(Y\) is a south pole.
  2. \(X\) is a south pole and \(Y\) is a north pole.
  3. Both \(X\) and \(Y\) are north poles.
  4. Both \(X\) and \(Y\) are south poles.
Show Answers Only

\(A\)

Show Worked Solution

→ Using the right hand rule, the palm faces into the page (direction of the force) and thumb points down the page (direction of the current).

→ The direction of the magnetic field is from \(X\) and \(Y\) (direction of fingers).

→ \(X\) is a north pole and \(Y\) is a south pole.

\(\Rightarrow A\)

Statistics, SPEC2 2023 VCAA 19 MC

A company accountant knows that the amount owed on any individual unpaid invoice is normally distributed with a mean of $800 and a standard deviation of $200.

What is the probability, correct to three decimal places, that in a random sample of 16 unpaid invoices the total amount owed is more than $13 500?

  1. 0.087
  2. 0.191
  3. 0.413
  4. 0.587
  5. 0.809
Show Answers Only

\(B\)

Show Worked Solution

\(n=16, \ E(X)=\mu=800 \)

\(X\ \sim\ N(800, 200^2)  \)

\(\bar{X} \sim N\Bigg{(}800, \dfrac{200^2}{\sqrt{16}}\Bigg{)} \sim N(800, 50^2) \)

\(\text{Pr} (\Sigma X>13\ 500) = \text{Pr}\Bigg{(}\bar{X} > \dfrac{13\ 500}{16}\Bigg{)} = \text{Pr}(\bar{X} > 843.75)= 0.190…\)

\(\Rightarrow B\)

Probability, MET2 2023 VCAA 15 MC

Let X be a normal random variable with mean of 100 and standard deviation of 20. Let Y be a normal random variable with mean of 80 and standard deviation of 10.

Which of the diagrams below best represents the probability density functions for X and Y, plotted on the same set of axes?
 

A.   
         B.   
         
C.   D.
         
E.
      
Show Answers Only

\(A\)

Show Worked Solution

\(X\sim N(100, 20^2)\ \ \ Y\sim N(80, 10^2)\)

\(\text{Mean }X>\text{Mean }Y \ \text{ie peak of }X\ \text{is to the right.}\)

\(\longrightarrow\ \text{Eliminate B and E}\)

\(\text{SD }X>\text{SD }Y\ \therefore\ X\ \text{has a greater spread ie curve flatter.} \)

\(\longrightarrow\ \text{Eliminate C and D}\)

 
\(\Rightarrow A\)

Vectors, SPEC2 2023 VCAA 17 MC

Consider the vectors  \(\underset{\sim}{\text{a}}=\alpha \underset{\sim}{\text{i}}+\underset{\sim}{\text{j}}-\underset{\sim}{\text{k}}, \ \underset{\sim}{\text{b}}=3 \underset{\sim}{\text{i}}+\beta \underset{\sim}{\text{j}}+4 \underset{\sim}{\text{k}}\)  and  \(\underset{\sim}{\text{c}}=2 \underset{\sim}{\text{i}}-7 \underset{\sim}{\text{j}}+\gamma \underset{\sim}{\text{k}}\), where \(\alpha, \beta, \gamma \in R\). If  \(\underset{\sim}{\text{a}} \times \underset{\sim}{\text{b}}=\underset{\sim}{\text{c}}\), then

  1. \(\alpha=-2, \ \beta=-1, \ \gamma=-5\)
  2. \(\alpha=-1, \ \beta=2, \ \gamma=-1\)
  3. \(\alpha=1, \ \beta=-2, \ \gamma=-5\)
  4. \(\alpha=-2, \ \beta=-1, \ \gamma=-1\)
  5. \(\alpha=1, \ \beta=-2, \ \gamma=5\)
Show Answers Only

\(C\)

Show Worked Solution

\begin {aligned}\underset{\sim}{\text{a}} \times \underset{\sim}{\text{b}}=\left|\begin{array}{ccc}
\underset{\sim}{\text{i}} & \underset{\sim}{\text{j}} & \underset{\sim}{\text{k}} \\ \alpha & 1 & -1 \\ 3 & \beta & 4\end{array}\right|\ & =(4+\beta)\underset{\sim}{\text{i}}-(4\alpha+3)\underset{\sim}{\text{j}}+(\alpha \beta-3) \underset{\sim}{\text{k}} \end{aligned}

\(\text{Equating coefficients given}\ \ \underset{\sim}{\text{a}} \times \underset{\sim}{\text{b}}=\underset{\sim}{\text{c}}:\)

\(4+\beta =2\ \ \Rightarrow \ \beta=-2 \)

\(4\alpha +3 = 7 \ \ \Rightarrow \ \alpha = 1 \)

\(\alpha \beta-3=\gamma \ \ \Rightarrow \ \gamma=-5 \)

\(\Rightarrow C\)

Mechanics, EXT2 M1 SM-Bank 5

A student throws a ball for his dog to retrieve. The position vector of the ball, relative to an origin \(O\) at ground level \(t\) seconds after release, is given by  \(  \underset{\sim}{\text{r}}{}_\text{B} (t)=5 t \underset{\sim}{\text{i}}+7 t \underset{\sim}{\text{j}}+(15 t-4.9 t^2+1.5) \underset{\sim}{\text{k}} \). Displacement components are measured in metres, where \(\underset{\sim}{\text{i}}\) is a unit vector to the east, \(\underset{\sim}{\text{j}}\) is a unit vector to the north and \( \underset{\sim} {\text{k}}\) is a unit vector vertically up.

Calculate the total vertical distance, in metres, travelled by the ball before it hits the ground. Give your answer correct to one decimal place.   (3 marks)

Show Answers Only

\(24.5\ \text{m} \)

Show Worked Solution

\(\text{Upwards distance}\ (z) = 15t-4.9t^2+1.5\)

\(\dfrac{dz}{dt}=15-9.8t\)

\(\text{Find}\ t\ \text{when}\ \dfrac{dz}{dt}=0\ \text{(vertical max):} \)

\(15-9.8t=0\ \ \Rightarrow \ \ t=\dfrac{15}{9.8} \)

\(z\Big{(}t=\frac{15}{9.8}\Big{)} = 15 \times \Big{(}\dfrac{15}{9.8}\Big{)}-4.9 \times \Big{(}\dfrac{15}{9.8}\Big{)}^2 + 1.5 \approx 12.98\ \text{m} \)
  

\(\text{At}\ \ t=0, \ z=1.5 \)

\(\therefore \text{Total vertical distance}\ = (12.98-1.5)+12.98 = 24.5\ \text{m (1 d.p.)} \)

Vectors, SPEC2 2023 VCAA 15 MC

If the sum of two unit vectors is a unit vector, then the magnitude of the difference of the two vectors is

  1. \(0\)
  2. \(\dfrac{1}{\sqrt{2}}\)
  3. \(\sqrt{2}\)
  4. \(\sqrt{3}\)
  5. \(\sqrt{5}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Vectors can be drawn as two sides of an equilateral triangle.}\)

\(\text{Using the cosine rule for the difference between the two vectors:}\)

\(c^2\) \(=a^2+b^2-2ab\, \cos C\)  
  \(=1+1-2\times -\dfrac{1}{2} \)  
  \(=3\)  
\(c\) \(=\sqrt{3}\)  

 
\(\Rightarrow D\)

Calculus, SPEC2 2023 VCAA 13 MC

A tourist in a hot air balloon, which is rising vertically at 2.5 m s\(^{-1}\), accidentally drops a phone over the side when the phone is 80 metres above the ground.

Assuming air resistance is negligible, how long in seconds, correct to two decimal places, does it take for the phone to hit the ground?

  1. 2.86
  2. 2.98
  3. 3.79
  4. 4.04
  5. 4.30
Show Answers Only

\(E\)

Show Worked Solution

\(\text{Take upward velocity as positive.}\)

\(u=2.5\ \text{ms}^{-1}, \ a=-9.8\ \text{ms}^{-2} \)

\(\text{Using}\ \ s=ut+\dfrac{1}{2}at^2,\)

\(\text{Find}\ t\ \text{when}\ \ s=-80\ \text{(by calc):} \)

\(-80=2.5t-4.9t^2\ \ \Rightarrow \ \ t=4.30\ \text{s}\)

\(\Rightarrow E\)

Mechanics, EXT2 M1 EQ-Bank 6

The acceleration, \(a\) ms\(^{-2}\), of a particle that starts from rest and moves in a straight line is described by  \(a=1+v\), where \(v\) ms\(^{-1}\) is its velocity after \(t\) seconds.

Determine the velocity of the particle after \( \log _e(e+1) \) seconds.   (3 marks)

Show Answers Only

\(e\ \text{ms}^{-1}\)

Show Worked Solution

\(\dfrac{dv}{dt}=1+v\ \ \Rightarrow \ \dfrac{dt}{dv} = \dfrac{1}{1+v} \)

\(t= \displaystyle{\int \dfrac{1}{1+v}\ dv} = \log_e{(1+v)}+c \)

\(\text{When}\ \ t=0, v=0\ \ \Rightarrow \ c=0 \)

\(t\) \(=\log_e(1+v) \)  
\(1+v\) \(=e^t\)  
\(v\) \(=e^t-1\)  

 
\(\text{At}\ \ t=\log_e(e+1): \)

\(v=e^{\log_e{(e+1)}}-1 = e+1-1=e\ \text{ms}^{-1} \)

Probability, MET2 2023 VCAA 10 MC

A continuous random variable \(X\) has the following probability density function
 

\(g(x) = \begin {cases}
\dfrac{x-1}{20}         &\ \ 1 \leq x < 6 \\
\\
\dfrac{9-x}{12}         &\ \ 6 \leq x < 9 \\
\\ 0 &\ \ \ \text{elsewhere}
\end{cases}\)
 

The value of \(k\) such that  \(\text{Pr}(X<k)=0.35\)  is

  1. \(\sqrt{14}-1\)
  2. \(\sqrt{14}+1\)
  3. \(\sqrt{15}-1\)
  4. \(\sqrt{15}+1\)
  5. \(1-\sqrt{15}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\displaystyle \int_{1}^{6} \dfrac{x-1}{20}\,dx=0.625>0.35\)

\(\therefore\ k\ \text{lies between 1 and 6 so solve for interval }1\leq x\leq k\)
 

\(\text{Solve for}\ k\ \text{(by CAS)}:\)

\(\displaystyle \int_{1}^{k} \dfrac{x-1}{20}\,dx=0.35\ \ \Rightarrow \ \ k=\sqrt{14}+1\)

\(\Rightarrow B\)

Calculus, MET2 2023 VCAA 7 MC

Let  \(f(x)=\log_{e}x\), where  \(x>0\)  and  \(g(x)=\sqrt{1-x}\), where  \(x<1\).

The domain of the derivative of \((f\circ g)(x)\) is

  1. \(x\in R\)
  2. \(x\in (-\infty, 1]\)
  3. \(x\in (-\infty, 1)\)
  4. \(x\in (0, \infty)\)
  5. \(x\in (0, 1)\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Given }f(x)=\log_{e}x\ \ \text{and}\ \ g(x)=\sqrt{1-x}\)

\((f\circ g)(x)=\log_{e}\sqrt{1-x}=\dfrac{1}{2}\log_{e}(1-x)\)

\((f\circ g)^{′}(x)=\dfrac{1}{2}\times\dfrac{-1}{1-x}=\dfrac{1}{2(x-1)}\ \text{  where}\ \ x<1\)

\(\Rightarrow C\)

Calculus, MET2 2023 VCAA 5 MC

Which one of the following functions has a horizontal tangent at \((0, 0)\)?

  1. \(y=x^{-\frac{1}{3}}\)
  2. \(y=x^{\frac{1}{3}}\)
  3. \(y=x^{\frac{2}{3}}\)
  4. \(y=x^{\frac{4}{3}}\)
  5. \(y=x^{\frac{3}{4}}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Index must be greater than 1 otherwise the gradient function }\)

\(\text{will not be defined at }x=0.\)

\(y=x^{\frac{4}{3}}\ \ \Rightarrow \ \dfrac{dy}{dx}=\frac{4}{3}x^{\frac{1}{3}}\ \ (\text{only option defined at}\ x=0)\)

\(\Rightarrow D\)

Algebra, MET2 2023 VCAA 4 MC

Consider the system of simultaneous equations below containing the parameter \(k\).

\(kx+5y\) \(=k+5\)
\(4x+(k+1)y\) \(=0\)

 
The value(s) of \(k\) for which the system of equations has infinite solutions are

  1. \(k\in \{-5, 4\}\)
  2. \(k\in \{-5\}\)
  3. \(k\in \{4\}\)
  4. \(k\in R\setminus \{-5, 4\}\)
  5. \(k\in R\setminus \{-5\}\)
Show Answers Only

\(B\)

Show Worked Solution
\(kx+5y\) \(=k+5\) \(\ \ \rightarrow\ \ \ \ \) \(y\) \(=-\dfrac{k}{5}x+\dfrac{k+5}{5}\)
\(4x+(k+1)y\) \(=0\) \(\ \ \rightarrow\ \ \ \ \) \(y\) \(=-\dfrac{4}{k+1}x+0\)

 
\(\text{Infinite solutions when gradients and }y\text{-intercepts equal.}\)
 
\(\text{Equating gradients (by CAS):}\)

\(-\dfrac{k}{5}=-\dfrac{4}{k+1}\ \ \Rightarrow\ \ k=4\ \text{ or}\ -5\)
   
\(\text{Equating intercepts:}\)

\(\dfrac{k+5}{5}=0\ \ \Rightarrow \ \ k=-5\)

\(k=-5\ \ \text{satisfies both equations (infinite solutions)}.\)

\(\Rightarrow B\)

Mean mark 55%.

Calculus, MET1 2022 VCAA 8

Part of the graph of `y=f(x)` is shown below. The rule `A(k)=k \ sin(k)` gives the area bounded by the graph of `f`, the horizontal axis and the line `x=k`.
 

  1. State the value of `A\left(\frac{\pi}{3}\right)`.   (1 mark)

  2. Evaluate `f\left(\frac{\pi}{3}\right)`.   (2 marks)

  3. Consider the average value of the function `f` over the interval `x \in[0, k]`, where `k \in[0,2]`.
  4. Find the value of `k` that results in the maximum average value.   (2 marks)

Show Answers Only

a.    ` (sqrt(3)pi)/6`

b.    `(3sqrt(3) +pi)/6`

c.    `k = pi/2`

Show Worked Solution
a.  `A(pi/3)` `= pi/3sin(pi/3)`  
  `= pi/3 xx sqrt(3)/2`  
  `= (sqrt(3)pi)/6`  
b.   `f(k)` `= A^{\prime}(k)`  
`f(k)` `=d/dx(k\ sin\ k)`  
  `= sin\ k + k\ cos\ k`  
`f(pi/3)` `= sin\ pi/3 + pi/3\ cos\ pi/3`  
  `= sqrt(3)/2 + pi/3 xx 1/2`  
  `= sqrt(3)/2 + pi/6`  
  `=(3sqrt(3) +pi)/6`  

♦♦♦ Mean mark (b) 25%.
MARKER’S COMMENT: Common error was giving derivative of `A(k)` as `k\ cos (k)`.
c.   Average value `= (A(k))/k`  
  `= 1/k(int_0^k f(x) d x)`  
  `= 1/k[x\ sin (x)]_0^k`  
  `= 1/k[k\ sin (k)] = sin\ k`  

 
`:.` Average value has a maximum value of 1 when `k = pi/2`


♦♦♦ Mean mark (c) 25%.
MARKER’S COMMENT: Students often chose unnecessarily complicated methods or inconsistently applied nomenclature when attempting this question.

Calculus, MET1 2022 VCAA 7

A tilemaker wants to make square tiles of size 20 cm × 20 cm.

The front surface of the tiles is to be painted with two different colours that meet the following conditions:

  • Condition 1 - Each colour covers half the front surface of a tile.
  • Condition 2 - The tiles can be lined up in a single horizontal row so that the colours form a continuous pattern.

An example is shown below.
 

There are two types of tiles: Type A and Type B.

For Type A, the colours on the tiles are divided using the rule `f(x)=4 \sin \left(\frac{\pi x}{10}\right)+a`, where `a \in R`.

The corners of each tile have the coordinates (0,0), (20,0), (20,20) and (0,20), as shown below.
 

  1.  i. Find the area of the front surface of each tile.   (1 mark)

    ii. Find the value of `a` so that a Type A tile meets Condition 1.   (1 mark)


Type B tiles, an example of which is shown below, are divided using the rule `g(x)=-\frac{1}{100} x^3+\frac{3}{10} x^2-2 x+10`.
 

  1. Show that a Type B tile meets Condition 1.   (3 marks)

  2. Determine the endpoints of `f(x)` and `g(x)` on each tile. Hence, use these values to confirm that Type A and Type B tiles can be placed in any order to produce a continuous pattern in order to meet Condition 2.   (2 marks)

Show Answers Only

a.i.    `400` cm²

a.ii.   `a = 10`

b.     `200` cm²

c.     See worked solution

Show Worked Solution
a.i   Area `= 20 xx 20`  
  `= 400` cm²  

  
a.ii  `a = 10`
 

b.   Area `=\int_0^{20} \frac{-x^3}{100}+\frac{3 x^2}{10}-2 x+10\ d x`  
  `=\left[\frac{-x^4}{400}+\frac{x^3}{10}-x^2+10 x\right]_0^{20}`  
  `=\left[-\frac{20^4}{400}+\frac{20^3}{10}-20^2+10 xx 20\right]-\left[0\right]`  
  `= – 400 +800 -400 +200`  
  `= 200` cm²  

 
`:.` The area of the coloured section of the Type B tile is 200 cm² which is half the 400 cm² area of the tile.
 

c.   
`f(0)` `=4 \sin \left(\frac{\pi(0)}{10}\right)+10 = 10`  
  `f(20)` `=4 \sin (2 \pi)+10 = 10`  
  `g(0)` `=\frac{-0}{100}+\frac{3(0)}{10}-2(0)+10=10`  
  `g(20)` `=-\frac{8000}{100}+\frac{200}{10}-2(20)+10 = 10`  

 
→`\ f(0) = f(20) = g(0) = g(20) =10`
 

`:.`    The endpoints for `f(x)` are `(0,10)` and `(20,10)` and for `g(x)` are also `(0,10)` and `(20,10)`.

So the tiles can be placed in any order to make the continuous pattern.


♦♦ Mean mark (c) 30%.
MARKER’S COMMENT: Students often only found `f(20)` and `g(20)`, however, `f(0)` and `g(0)` also needed to be found to verify the pattern match.

Graphs, MET1 2022 VCAA 6

The graph of `y=f(x)`, where `f:[0,2 \pi] \rightarrow R, f(x)=2 \sin(2x)-1`, is shown below.
 

  1. On the axes above, draw the graph of `y=g(x)`, where `g(x)` is the reflection of `f(x)` in the horizontal axis.   (2 marks)
  2. Find all values of `k` such that `f(k)=0` and `k \in[0,2 \pi]`.   (3 marks)
  3. Let `h: D \rightarrow R, h(x)=2 \sin(2x)-1`, where `h(x)` has the same rule as `f(x)` with a different domain.
  4. The graph of `y=h(x)` is translated `a` units in the positive horizontal direction and `b` units in the positive vertical direction so that it is mapped onto the graph of `y=g(x)`, where `a, b \in(0, \infty)`.
    1. Find the value for `b`.   (1 mark)

    2. Find the smallest positive value for `a`.   (1 mark)

    3. Hence, or otherwise, state the domain, `D`, of `h(x)`.   (1 mark)

Show Answers Only

a.    Graph `y=g(x)`

b.    `\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{13 \pi}{12}, \frac{17 \pi}{12}`

c.i.    `b=2`

cii.    `=\frac{\pi}{2}`

ciii.   `\left[-\frac{\pi}{2}, \frac{3 \pi}{2}\right]`

Show Worked Solution

a.

b.  `2 \sin (2 k)-1` `=0`       `0<=k<=2\pi`  
`sin (2 k)` `=1/2`  
`2k` `=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{13 \pi}{6}, \frac{17 \pi}{6}`  
`k` `=\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{13 \pi}{12}, \frac{17 \pi}{12}`  

 

c.i   ` 2 \sin 2(x-a)-1+b` `=-(2 \sin 2 x-1)`  
`:.\ -1+b` `=1`  
`b` `=2`  

 

c.ii  `2 \sin 2(x-a)` `=-(2 \sin 2 x-1)`  
`\sin (2 x-2 a)` `=-\sin 2 x`  
`\therefore 2 a` `=\pi`  
`a` `=\frac{\pi}{2}`  

♦ Mean mark (c.ii) 50%.
MARKER’S COMMENT: Students confused vertical and horizontal translations. Common error `a=\frac{\pi}{4}.`

c.iii The domain for `f(x)` is `[0,2pi]`

`:. \ D` is `\left[-\frac{\pi}{2}, \frac{3 \pi}{2}\right]`

 
♦♦♦ Mean mark (c.iii) 10%.
MARKER’S COMMENT: Common error was translating in the wrong direction. A common incorrect answer was `\left[-\frac{\pi}{2}, \frac{5 \pi}{2}\right].`

Probability, MET1 2022 VCAA 4

A card is drawn from a deck of red and blue cards. After verifying the colour, the card is replaced in the deck. This is performed four times.

Each card has a probability of `\frac{1}{2}` of being red and a probability of `\frac{1}{2}` of being blue.

The colour of any drawn card is independent of the colour of any other drawn card.

Let `X` be a random variable describing the number of blue cards drawn from the deck, in any order.

  1. Complete the table below by giving the probability of each outcome.   (2 marks)
     

  1. Given that the first card drawn is blue, find the probability that exactly two of the next three cards drawn will be red.  (1 mark)

  2. The deck is changed so that the probability of a card being red is `\frac{2}{3}` and the probability of a card being blue is `\frac{1}{3}`.
  3. Given that the first card drawn is blue, find the probability that exactly two of the next three cards drawn will be red.   (2 marks)

Show Answers Only

a.   Complete table

\begin{array} {|c|c|c|c|c|c|}
\hline x & 0 & 1 & 2 & 3 & 4 \\
\hline {Pr}(X=x) & 1/16 & 4/16 & 6/16 & 4/16 & 1/16 \\
\hline \end{array}

b.    `3/8`

c.    `4/9`

Show Worked Solution

a.   Using binomial distribution where `n=4` and  `p= 1/2`

`text{Pr}(X=1)` `=\ ^4 C_1*(1/2)^1*(1/2)^3 = 4*1/2*1/8` `=4/16`
`text{Pr}(X=3)` `=\ ^4 C_3*(1/2)^3*(1/2)^1= 4*1/8*1/2` `=4/16`
`text{Pr}(X=4)` `=\ ^4 C_4*(1/2)^4*(1/2)^0= 1*1/16*1` `=1/16`

\begin{array} {|c|c|c|c|c|c|}
\hline x & 0 & 1 & 2 & 3 & 4 \\
\hline {Pr}(X=x) & 1/16 & 4/16 & 6/16 & 4/16 & 1/16 \\
\hline \end{array}

b.   Trials are independent of first trial.

Binomial where `n=3, p=1/2`

`text{Pr}(X=2) = \ ^3 C_2*(1/2)^2*(1/2)^1 = 3*1/8 = 3/8`


♦♦ Mean mark (b) 40%.

c.   Trials are independent.

Binomial where `n=3, p=2/3`

`text{Pr}(X=2) = \ ^3 C_2*(2/3)^2*(1/3)^1 = 3*4/9*1/3 = 4/9`


♦ Mean mark 55%.
MARKER’S COMMENT: Many students wrongly treated this question as conditional probability.

Calculus, SPEC2 2023 VCAA 11 MC

The area of the curved surface generated by revolving part of the curve with equation  \(y=\cos ^{-1}(x)\)  from  \((0, \dfrac{\pi}{2})\) to \((1,0)\) about the \(y\)-axis can be found by evaluating

  1. \(2 \pi \displaystyle {\int_0^{\dfrac{\pi}{2}}\left(\cos ^{-1}(x) \sqrt{1+\dfrac{1}{x^2-1}}\right)} d x\)
  2. \(2 \pi \displaystyle {\int_0^1\left(\cos ^{-1}(x) \sqrt{1+\dfrac{1}{x^2-1}}\right)} d x\)
  3. \(2 \pi \displaystyle {\int_0^{\dfrac{\pi}{2}} \cos (y) \sqrt{1-\sin ^2(y)}} d y\)
  4. \(2 \pi \displaystyle {\int_0^{\dfrac{\pi}{2}} \sqrt{1+u^2} \ d u}\), where \(u=\sin (y)\)
  5. \(2 \pi \displaystyle {\int_0^1 \sqrt{1+u^2} \ d u } \), where \(u=\sin (y)\)
Show Answers Only

\(E\)

Show Worked Solution

\(I = 2 \pi \displaystyle {\int_0^{\dfrac{\pi}{2}} \cos (y) \sqrt{1+\sin ^2(y)}}\ d y\)

\(\text{Let}\ \ u=\sin(y) \)

\(\dfrac{du}{dy}=\cos(y)\ \ \Rightarrow \ \ du=\cos(y)\ dy \)

\(\text{When}\ \ y=\dfrac{\pi}{2}\ \Rightarrow \ u=1,\ \ y=0\ \Rightarrow u=0 \)

\(\therefore I = 2 \pi \displaystyle {\int_0^{1} \sqrt{1+u^2}}\ d u\)

\(\Rightarrow E\)

Calculus, SPEC2 2023 VCAA 10 MC

If  \(I_n=\displaystyle {\int_0^1\left((1-x)^n e^x\right) d x}\), where  \(n \in N\), then for  \(n \geq 1, I_n\) equals

  1. \(-1+n I_{n-1}\)
  2. \(n I_{n-1}\)
  3. \(-1-n I_{n-1}\)
  4. \(-n I_{n-1}\)
  5. \((1-x)^n e^x+n I_{n-1}\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Using integration by parts:}\)

\begin{aligned}
u &=(1-x)^n, & u^{′} &=n(1-x)^{n-1}\\
v^{′} & = e^x, & v & = e^x\  \\
\end{aligned}

\begin{aligned}
\int_0^1 ((1-x)^n e^x)\,dx\ &= uv-\int u^{′} v \ dx\\
& = \Bigr[(1-x)^n\,e^x\Bigr]_0 ^1- \int_0 ^1 n(1-x)^{n-1}\,e^x\ dx  \\
& =-1-n\int_0 ^1 (1-x)^{n-1}\,e^x\ dx  \\
& =-1-nI_{n-1}  \\
\end{aligned}

\(\Rightarrow A\)

Calculus, SPEC2 2023 VCAA 8 MC

Initially a spa pool is filled with 8000 litres of water that contains a quantity of dissolved chemical. It is discovered that too much chemical is contained in the spa pool water. To correct this situation, 20 litres of well-mixed spa pool water is pumped out every minute while 15 litres of fresh water is pumped in each minute.

Let \(Q\) be the number of kilograms of chemical that remains dissolved in the spa pool after \(t\) minutes. The differential equation relating \(Q\) to t is

  1. \(\dfrac{d Q}{d t}=\dfrac{4 Q}{t-1600}\)
  2. \(\dfrac{d Q}{d t}=\dfrac{-Q}{400}\)
  3. \(\dfrac{d Q}{d t}=\dfrac{3 Q}{t-1600}\)
  4. \(\dfrac{d Q}{d t}=\dfrac{3 Q}{1600-t}\)
  5. \(\dfrac{d Q}{d t}=\dfrac{4 Q}{1600-t}\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Volume}\ = 8000-5t \)

\(Q(t) = \dfrac{Q}{8000-5t} \)

\(\dfrac{dQ}{dt}=-\dfrac{20Q}{8000-5t} = \dfrac{4Q}{t-1600} \)

\(\Rightarrow A\)

Calculus, SPEC2 2023 VCAA 7 MC

The direction field for a differential equation is shown above. On a certain solution curve of this differential equation, \(y=2\)  when  \(x=-1\).

The value of \(y\) on the same solution curve when  \(x=1.5\)  is closest to

  1. \(-0.5\)
  2. \(0\)
  3. \(0.5\)
  4. \(1.0\)
  5. \(1.5\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Trace a curve that does not cross slope lines from}\ (-1,2).\)

\(\text{Curve crosses}\ \ x=1\ \ \text{at}\ \ y\approx 1.\)

\(\Rightarrow D\)

Calculus, SPEC2 2023 VCAA 6 MC

Consider the following pseudocode.

define \(f(x, y)=e^{x y}\)

\(\begin{aligned} & x \leftarrow 0 \\ & y \leftarrow 0 \\ & h \leftarrow 0.5 \\ & n \leftarrow 0\end{aligned}\)

while \(n \geq 0\)

\(\begin{aligned} & y \leftarrow y+h \times f(x, y) \\ & x \leftarrow x+h \\ & n \leftarrow n+1\end{aligned}\)

print \(y\)

end while

After how many iterations will the pseudocode print 2.709 ?

  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
Show Answers Only

\(C\)

Show Worked Solution

\(\text{1st iteration:}\ 0+\dfrac{1}{2} \times e^0=\dfrac{1}{2} \)

\(\text{2nd iteration:}\ \dfrac{1}{2} +\dfrac{1}{2} \times e^{\frac{1}{4}}=1.142… \)

\(\text{3rd iteration:}\ 1.142 +\dfrac{1}{2} \times e^{1 \times 1.142}=2.7085… \)

\(\Rightarrow C\)

Complex Numbers, SPEC2 2023 VCAA 5 MC

Let \(z\) be a complex number where  \(\operatorname{Re}(z)>0\)  and  \(\operatorname{Im}(z)>0\).

Given  \(|\bar{z}|=4\)  and  \(\arg \left(z^3\right)=-\pi\), then \(z^2\) is equivalent to

  1. \( {4z} \)
  2. \( -2 \bar{z} \)
  3. \( 3z \)
  4. \(\bar{z}^2\)
  5. \(-4 \bar{z}\)
Show Answers Only

\(E\)

Show Worked Solution

\(\arg(z^3) = -\pi=\pi\ \ \Rightarrow \ \arg(z)=\dfrac{\pi}{3}\)

\(z\) \(=4\text{cis}\Big{(}\dfrac{\pi}{3}\Big{)}\ \ (\abs{z}=\abs{\bar z})\)  
\(z^2\) \(=16\text{cis}\Big{(}\dfrac{2\pi}{3}\Big{)} \)  
  \(=-4 \times 4\text{cis}\Big{(}-\dfrac{\pi}{3}\Big{)} \)  
  \(=-4\,\bar z\)  

 
\(\Rightarrow E\)

Calculus, MET1 2023 VCAA 7

Consider \(f:(-\infty, 1]\rightarrow R, f(x)=x^2-2x\). Part of the graph of  \(y=f(x)\)  is shown below.
 

  1. State the range of \(f\).   (1 mark)
  2. Sketch the graph of the inverse function  \(y=f^{-1}(x)\) on the axes above. Label any endpoints and axial intercepts with their coordinates.   (2 marks)
  3. Determine the equation of the domain for the inverse function  \(f^{-1}\).   (2 marks)
  4. Calculate the area of the regions enclosed by the curves of \(f,\ f^{-1}\)  and  \(y=-x\).   (2 marks)
Show Answers Only

a.    \([-1, \infty)\)

b.   

c.    \(f^{-1}(x)=1-\sqrt{x+1}\)

\(\text{Domain}\ [-1, \infty)\)

d.    \(A=\dfrac{1}{3}\)

Show Worked Solution

a.    \([-1, \infty)\)

b.   

c.    \(\text{When }f(x)\ \text{is written in turning point form}\)

\(y=(x-1)^2-1\)
 

\(\text{Inverse function: swap}\ x \leftrightarrow y\)

\(x\) \(=(y-1)^2-1\)
\(x+1\) \(=(y-1)^2\)
\(-\sqrt{x+1}\) \(=y-1\)
\(f^{-1}(x)\) \(=1-\sqrt{x+1}\)

 
\(\text{Domain}\ [-1, \infty)\)


♦ Mean mark (c) 48%.
MARKER’S COMMENT: Common error → writing the function as \(f^{-1}(x)=1+\sqrt{x+1}\).

d.     \(\text{One strategy of many possibilities:}\)

  \(A\) \(=2\displaystyle \int_{0}^{1} \big(-x-(x^2-2x)\big)\,dx\)
    \(=2\displaystyle \int_{0}^{1} \big(x-x^2\big)\,dx\)
    \(=2\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1\)
    \(=2\bigg(\dfrac{1}{2}-\dfrac{1}{3}-(0)\bigg)\)
    \(=\dfrac{1}{3}\)

♦♦♦ Mean mark (d) 24%.
MARKER’S COMMENT: Using the symmetry properties of the graph and its inverse helped answer this question efficiently.

Calculus, MET1 2023 VCAA 5

  1. Evaluate  \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx\).   (1 mark)
  2. Hence, or otherwise, find all values of \(k\) such that \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx=\displaystyle \int_{0}^{\frac{\pi}{2}} \cos(x)\,dx\), where \(-3\pi<k<2\pi\).   (3 marks)
Show Answers Only

a.    \(\dfrac{1}{2}\)

b.    \(k=\dfrac{-11\pi}{6},\ \dfrac{-7\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)

Show Worked Solution
a.    \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx\) \(=\left[-\cos x\right]_0^\frac{\pi}{3}\)
    \(=-\cos\dfrac{\pi}{3}+\cos 0\)
    \(=-\dfrac{1}{2}+1\)
    \(=\dfrac{1}{2}\)

 

b.    \(\displaystyle \int_{k}^{\frac{\pi}{2}} \cos(x)\,dx\) \(=\left[\sin x\right]_k^\frac{\pi}{2}\)
    \(=\sin\bigg(\dfrac{\pi}{2}\bigg)-\sin (k)\)
    \(=1-\sin (k)\)

 
\(\text{Using part (a):}\)

\(1-\sin (k)\) \(=\dfrac{1}{2}\)
\(\sin (k)\) \(=\dfrac{1}{2}\)
\(\therefore\ k\) \(=\dfrac{-11\pi}{6},\ \dfrac{-7\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)

Calculus, MET1 2023 VCAA 4

The graph of  \(y=x+\dfrac{1}{x}\) is shown over part of its domain.
 

Use two trapeziums of equal width to approximate the area between the curve, the \(x\)-axis and the lines  \(x=1\)  and  \(x=3\).   (2 marks)

Show Answers Only

\(5\dfrac{1}{6}\)

Show Worked Solution

\(\text{Trapezium rule approximation (see formula sheet):}\)

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} x\rule[-1ex]{0pt}{0pt} & 1&2&3 \\
\hline
\rule{0pt}{2.5ex} f(x)\rule[-1ex]{0pt}{0pt} & 1+1=2 & 2+\dfrac{1}{2}=\dfrac{5}{2} & 3+ \dfrac{1}{3}=\dfrac{10}{3}\\
\hline
\end{array}

\(\text{Area}\) \(\approx \dfrac{3-1}{2\times 2}\Bigg[2+2\times\dfrac{5}{2}+\dfrac{10}{3}\Bigg]\)
  \(\approx\dfrac{1}{2}\Bigg[\dfrac{6}{3}+\dfrac{15}{3}+\dfrac{10}{3}\Bigg]\)
  \(\approx5\dfrac{1}{6}\)

Calculus, MET1 2022 VCAA 1b

Find and simplify the rule of `f^{\prime}(x)`, where `f:R \rightarrow R, f(x)=\frac{\cos (x)}{e^x}`.   (2 marks)

Show Answers Only

`\frac{-(\sin x+\cos x)}{e^x}`

Show Worked Solution

 Using the quotient rule

`f(x)` `=\frac{\cos (x)}{e^x}`  
`f^{\prime}(x)` `=\frac{-e^x \sin x-e^x \cos x}{e^{2 x}}`  
  `= \frac{-e^x(\sin x+\cos x)}{e^{2 x}}`  
  `=\frac{-(\sin x+\cos x)}{e^x}`  

Algebra, MET1 2023 VCAA 2

Solve  \(e^{2x}-12=4e^{x}\)  for  \(x\ \in\ R\).   (3 marks)

Show Answers Only

\(x=\log_{e}6\)

Show Worked Solution
\(e^{2x}-12\) \(=4e^{x}\)
\(e^{2x}-4e^{x}-12\) \(=0\)

 
\(\text{Let}\ \ u=e^{x}:\)

\(u^2-4u-12\) \(=0\)
\((u-6)(u+2)\) \(=0\)
 

\(\Rightarrow u=6\ \ \ \text{or}\ -2\)

\(\therefore e^{x}\) \(=6\ \ \ \ \ \ \ \ \ \ \text{or}\ \ \ \ \ \ e^{x}=-2\ \text{(no solution)}\)  
\(x\) \(=\log_{e}6 \)  

Calculus, MET1 2023 VCAA 1a

Let  \(y=\dfrac{x^2-x}{e^x}\).

Find and simplify \(\dfrac{dy}{dx}\).   (2 marks)

Show Answers Only

\(-\Bigg(\dfrac{x^2-3x+1}{e^x}\Bigg) \)

Show Worked Solution

\(\text{Using the quotient rule:}\)

\(\dfrac{dy}{dx}\) \(=\dfrac{e^x(2x-1)-(x^2-x)e^x}{(e^x)^2}\)
  \(=\dfrac{e^x(-x^2+3x-1)}{e^{2x}}\)
  \(=\dfrac{-x^2+3x-1}{e^x}\)

Trigonometry, SPEC2 2023 VCAA 3 MC

In the interval  \(-\pi \leq x \leq \pi\), the graph of  \(y=a+\sec (x)\), where  \(a \in R\), has two \(x\)-intercepts when

  1. \(0 \leq a \leq 1\)
  2. \(-1<a<1\)
  3. \(a \leq-1\)  or  \(a>1\)
  4. \(-1 \leq a<0\)
  5. \(a<-1\)  or  \(a \geq 1\)
Show Answers Only

\(E\)

Show Worked Solution

\(\text{Sketch}\ \ y=\sec(x)\ \ \text{for}\ \  -\pi \leq x \leq \pi \)

\(\text{Translate vertically to test for values of}\ a\ \text{that satisfy}\)

\(\text{two intercepts on the}\ x\text{-axis.}\)

\(a \geq 1\ \ \text{or}\ \ a \lt -1\)

\(\Rightarrow E\)

Vectors, SPEC1 2023 VCAA 10

The position vector of a particle at time \(t\) seconds is given by

\(\underset{\sim}{\text{r}}(t)=\big{(}5-6 \ \sin ^2(t) \big{)} \underset{\sim}{\text{i}}+(1+6 \ \sin (t) \cos (t)) \underset{\sim}{\text{j}}\), where \(t \geq 0\).

  1. Write \(5-6\, \sin ^2(t)\) in the form \(\alpha+\beta\, \cos (2 t)\), where \(\alpha, \beta \in Z^{+}\).  (1 mark)
  1. Show that the Cartesian equation of the path of the particle is \((x-2)^2+(y-1)^2=9.\)  (2 marks)
  1. The particle is at point \(A\) when \(t=0\) and at point \(B\) when \(t=a\), where \(a\) is a positive real constant.
  2. If the distance travelled along the curve from \(A\) to \(B\) is \(\dfrac{3 \pi}{4}\), find \(a\).   (1 mark)
  1. Find all values of \(t\) for which the position vector of the particle, \(\underset{\sim}{\text{r}}(t)\), is perpendicular to its velocity vector, \(\underset{\sim}{\dot{\text{r}}}(t)\).   (2 marks)
Show Answers Only

a.    \(2+3\, \cos (2 t) \)

b.   \( x=2+3\, \cos (2 t) \Rightarrow \dfrac{x-2}{3}=\cos (2 t)\)

\(y=1+6\, \sin (t) \cos (t)=1+3\, \sin (2 t) \Rightarrow \dfrac{y-1}{3}=\sin (2 t)\)

\begin{aligned}
\sin ^2(2 t)+\cos ^2(2 t) &=1 \\
\left(\dfrac{x-2}{3}\right)^2+\left(\dfrac{y-1}{3}\right)^2 & =1 \\
(x-2)^2+(y-1)^2 &=9
\end{aligned}

c.    \(a=\dfrac{\pi}{8}\)

d.    \(t =\dfrac{1}{2} \tan ^{-1}\left(\dfrac{1}{2}\right)+\dfrac{k \pi}{2}\ \ (\text{where}\ k=0,1,2,…) \)

Show Worked Solution

a.  \(5-6\, \sin ^2(t)=5-6 \times \dfrac{1}{2}(1-\cos (2 t))\)

\(\ \ \ \quad \quad \quad \quad \quad \quad \begin{aligned}
& =5-3+3\, \cos (2 t) \\
& =2+3\, \cos (2 t)
\end{aligned}\)
 

b.   \( x=2+3\, \cos (2 t) \Rightarrow \dfrac{x-2}{3}=\cos (2 t)\)

\(y=1+6\, \sin (t) \cos (t)=1+3\, \sin (2 t) \Rightarrow \dfrac{y-1}{3}=\sin (2 t)\)

\begin{aligned}
\sin ^2(2 t)+\cos ^2(2 t) &=1 \\
\left(\dfrac{x-2}{3}\right)^2+\left(\dfrac{y-1}{3}\right)^2 & =1 \\
(x-2)^2+(y-1)^2 &=9
\end{aligned}

 
c.  \(\text { Motion is circular, centre }(2,1) \text {, radius }=3\)

\begin{aligned}
\text { Arc length } & = r \theta \\
3 \theta & =\dfrac{3 \pi}{4} \\
\theta & =\dfrac{\pi}{4}
\end{aligned}

\(\therefore a=\dfrac{\pi}{8}\)
 

d.    \(\underset{\sim}{r}=(2+3\, \cos (2 t)) \underset{\sim}{i}+(1+3\, \sin (2 t)) \underset{\sim}{j}\)

\(\underset{\sim}{\dot{r}}=-6\, \sin (2 t) \underset{\sim}{i}+6\, \cos (2 t) \underset{\sim}{j}\)

\(\text { Find } t \text { when } \underset{\sim}{r} \cdot \underset{\sim}{\dot{r}}=0 \text { : }\)

\begin{aligned}
\underset{\sim}{r} \cdot \underset{\sim}{\dot{r}} & =6\left(\begin{array}{l}
2+3\, \cos (2 t) \\
1+3\, \sin (2 t)
\end{array}\right)\left(\begin{array}{l}
-\sin (2 t) \\
\cos (2 t)
\end{array}\right) \\
0 &=-2\,\sin (2 t)-3\, \cos (2 t) \sin (2 t)+\cos (2 t)+3\, \cos (2 t) \sin (2 t)\\
0 &=-2\, \sin (2 t)+\cos (2 t)\\
2\, \sin (2 t) &=\cos (2 t)\\
\tan (2 t) & =\dfrac{1}{2} \\
2 t & =\tan ^{-1}\left(\dfrac{1}{2}\right)+k \pi \\
t & =\dfrac{1}{2} \tan ^{-1}\left(\dfrac{1}{2}\right)+\dfrac{k \pi}{2}\ \ (\text{where}\ k=0,1,2,…) 
\end{aligned}

Functions, SPEC2 2023 VCAA 2 MC

The graph of  \(y=\dfrac{x^3}{a x^2+b x+c}\)  has asymptotes given by  \(y=2 x+1\)  and  \(x=1\). The values of \(a, b\) and \(c\) are, respectively

  1. \(2,-4,2\)
  2. \( \dfrac{1}{2},-\dfrac{1}{4},-\dfrac{1}{4} \)
  3. \( \dfrac{1}{2}, \dfrac{1}{4},-\dfrac{3}{4} \)
  4. \( \dfrac{1}{2},-\dfrac{1}{4},-\dfrac{3}{4} \)
  5. \(2,-4,-8\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Using partial fractions (by calc):}\)

\(\text{Remainder}\ =  \dfrac{x}{a}-\dfrac{b}{a^2} =  2x+1\)

\(a=\dfrac{1}{2},\ \ b=-\dfrac{1}{4} \)

\(x=1\ \ \text{is a solution of}\ \ ax^2+bx+c=0\)

\(c=-\dfrac{1}{4} \)

\(\Rightarrow B\)

Vectors, SPEC1 2023 VCAA 9

A plane contains the points \( A(1,3,-2), B(-1,-2,4)\) and \( C(a,-1,5)\), where \(a\) is a real constant. The plane has a \(y\)-axis intercept of 2 at the point \(D\).

  1. Write down the coordinates of point \(D\).   (1 mark)
  1. Show that \(\overrightarrow{A B}\) and \(\overrightarrow{A D}\) are \(-2 \underset{\sim}{\text{i}}-5 \underset{\sim}{\text{j}}+6 \underset{\sim}{\text{k}}\)  and  \(-\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}+2 \underset{\sim}{\text{k}}\), respectively.   (1 mark)
  1. Hence find the equation of the plane in Cartesian form.  (2 marks)
  1.  Find \(a\).   (1 mark)
  1.  \(\overline{A B}\) and \(\overline{A D}\) are adjacent sides of a parallelogram. Find the area of this parallelogram.   (1 mark)
Show Answers Only

a.    \(D(0,2,0)\)

b.    \(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\left(\begin{array}{c}
-1 \\
-2 \\
4
\end{array}\right)-\left(\begin{array}{c}
1 \\
3 \\
-2
\end{array}\right)=\left(\begin{array}{c}
-2 \\
-5 \\
6
\end{array}\right)=-2 \underset{\sim}{\text{i}}-5\underset{\sim}{\text{j}}+6 \underset{\sim}{\text{k}}\)

\(\overrightarrow{A D}=\overrightarrow{O D}-\overrightarrow{O A}=\left(\begin{array}{c}0 \\ 2 \\ 0\end{array}\right)-\left(\begin{array}{c}1 \\ 3 \\ -2\end{array}\right)=\left(\begin{array}{c}-1 \\ -1 \\ 2\end{array}\right)=-\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}+2\underset{\sim}{\text{k}}\)

c.    \(4 x+2 y+3 z =4\)

d.    \(a=-\dfrac{9}{4}\)

e.    \(A=\sqrt{29}\)

Show Worked Solution

a.    \(D(0,2,0)\)

 
b.    \(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\left(\begin{array}{c}
-1 \\
-2 \\
4
\end{array}\right)-\left(\begin{array}{c}
1 \\
3 \\
-2
\end{array}\right)=\left(\begin{array}{c}
-2 \\
-5 \\
6
\end{array}\right)=-2 \underset{\sim}{\text{i}}-5\underset{\sim}{\text{j}}+6 \underset{\sim}{\text{k}}\)

\(\overrightarrow{A D}=\overrightarrow{O D}-\overrightarrow{O A}=\left(\begin{array}{c}0 \\ 2 \\ 0\end{array}\right)-\left(\begin{array}{c}1 \\ 3 \\ -2\end{array}\right)=\left(\begin{array}{c}-1 \\ -1 \\ 2\end{array}\right)=-\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}+2\underset{\sim}{\text{k}}\)

 
c.    \(\overrightarrow{A B} \times \overrightarrow{A D}=\left|\begin{array}{ccc}
\underset{\sim}{\text{i}} & \underset{\sim}{\text{j}} & \underset{\sim}{\text{k}} \\ -2 & -5 & 6 \\ -1 & -1 & 2\end{array}\right|\)

\(\ \quad \quad \quad \quad \   \begin{aligned} & =(-10+6)\underset{\sim}{\text{i}}-(-4+6)\underset{\sim}{\text{j}}+(2-5) \underset{\sim}{\text{k}} \\ & =-4 \underset{\sim}{i}-2 \underset{\sim}{j}-3 \underset{\sim}{k}\end{aligned}\)

\(\text{Plane:}\ \ -4 x-2 y-3 z=k\)

\(\text{Substitute}\  D(0,2,0)\  \text{into equation}\ \ \Rightarrow \ \text{k}=-4\)

\begin{aligned}
\therefore-4 x-2 y-3 z & =-4 \\
4 x+2 y+3 z & =4
\end{aligned}

 
d.    \(\text{Find}\ a\ \Rightarrow \ \text{substitute}\ (a, 1,-5)\ \text{into plane equation:}\)

\begin{aligned}
-4 & =-4a+2-15 \\
4 a & =-9 \\
a & =-\dfrac{9}{4}
\end{aligned}

e.     \(\text{Area}\) \(=|\overrightarrow{A B} \times \overrightarrow{A D}|\)
    \(=|-4\underset{\sim}{i}-2\underset{\sim}{j}-3\underset{\sim}{k}|\)
    \(=\sqrt{16+4+9} \)
    \(=\sqrt{29}\)

Functions, SPEC1 2023 VCAA 1

Consider the function \(f\) with rule  \(f(x)=\dfrac{x^2+x-6}{x-1}\).

  1. Show that the rule for the function \(f\) can be written as  \(f(x)=x+2-\dfrac{4}{x-1}\).   (1 mark)
  1. Sketch the graph of \(f\) on the axes below, labelling any asymptotes with their equations.   (3 marks)

   

Show Answers Only
a.     \(f(x)\) \(=\dfrac{x^2+x-6}{x-1}\)
    \(=\dfrac{(x-1)(x+2)-4}{x-1}\)
    \(=x+2-\dfrac{4}{x-1} \)

 
b.    \(\text{Asymptotes at:}\ x=1, \ y=x+2 \)

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -3 & -1 & 0 & 1 & 2 \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 0 & 3 & 6 & \infty & 0 \\
\hline
\end{array}

 

Show Worked Solution
a.     \(f(x)\) \(=\dfrac{x^2+x-6}{x-1}\)
    \(=\dfrac{(x-1)(x+2)-4}{x-1}\)
    \(=x+2-\dfrac{4}{x-1} \)

 
b.    \(\text{Asymptotes at:}\ x=1, \ y=x+2 \)

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -3 & -1 & 0 & 1 & 2 \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 0 & 3 & 6 & \infty & 0 \\
\hline
\end{array}

 

Statistics, SPEC1 2023 VCAA 6

Josie travels from home to work in the city. She drives a car to a train station, waits, and then rides on a train to the city. The time, \(X_c \) minutes, taken to drive to the station is normally distributed with a mean of 20 minutes \( (\mu_c=20) \) and standard deviation of 6 minutes \((\sigma_c=6) \). The waiting time, \( X_w \) minutes, for a train is normally distributed with a mean of 8 minutes \( (\mu_w=8) \) and standard deviation of \( \sqrt{3} \) minutes \( (\sigma_w=\sqrt{3}) \). The time, \( X_t \) minutes, taken to ride on a train to the city is also normally distributed with a mean of 12 minutes \( (\mu_t=12) \) and standard deviation of 5 minutes \( (\sigma_t=5) \). The three times are independent of each other.

  1. Find the mean and standard deviation of the total time, in minutes, it takes for Josie to travel from home to the city.   (2 marks)
  1. Josie's waiting time for a train on each work day is independent of her waiting time for a train on any other work day. The probability that, for 12 randomly chosen work days, Josie's average waiting time is between 7 minutes 45 seconds and 8 minutes 30 seconds is equivalent to \( \text{Pr}(a<Z<b)\), where \(Z \sim \text{N}(0,1)\) and \(a\) and \(b\) are real numbers.
  2. Find the values of \(a\) and \(b\).   (2 marks)
Show Answers Only

a.    \(E[X_c+X_w+X_t]=40\)

\(\sigma[X_c+X_w+X_t]=8\)

b.    \(a=-0.5, \ b=1\)

Show Worked Solution

a.    \(X_c ∼ N(20, 6^2), X_w ∼ N(8, (\sqrt{3})^2), X_t ∼ N(12, 5^2) \)

\(E[X_c+X_w+X_t]\) \(=E[X_c]+E[X_w]+E[X_t]\)  
  \(=20+8+12\)  
  \(=40\)  

 

\(\text{Var}[X_c+X_w+X_t]\) \(=\text{Var}[X_c]+\text{Var}[X_w]+\text{Var}[X_t]\)  
  \(=36+3+25\)  
  \(=64\)  
\(\sigma[X_c+X_w+X_t]\) \(=8\)  

 
b.    \(E(\bar X)=\mu_w=8,\ \ \sigma(\bar X)=\dfrac{\sigma_w}{\sqrt{12}}=\dfrac{\sqrt3}{\sqrt{12}}=\dfrac{1}{2} \)

\(\text{Pr}(7.75<\bar X<8.5)\) \(=\text{Pr} \Bigg{(} \dfrac{7.75-8}{\frac{1}{2}}<Z<\dfrac{8.5-8}{\frac{1}{2}}\Bigg{)} \)  
  \(=\text{Pr} \Bigg{(} \dfrac{-0.25}{\frac{1}{2}}<Z<\dfrac{0.5}{\frac{1}{2}}\Bigg{)} \)  
  \(=\text{Pr}(-0.5<Z<1)\)  

 
\(\therefore a=-0.5, \ b=1\)

Calculus, SPEC1 2023 VCAA 4

Consider the relation \(x\, \arcsin \left(y^2\right)=\pi\).

Use implicit differentiation to find \(\dfrac{d y}{d x}\) at the point \(\left(6, \dfrac{1}{\sqrt{2}}\right)\).

Give your answer in the form \(-\dfrac{\pi \sqrt{a}}{b}\), where  \(a, b \in Z^{+}\).   (3 marks)

Show Answers Only

\(\dfrac{dy}{dx}=-\dfrac{\pi \sqrt6}{144} \)

Show Worked Solution
\(\dfrac{d}{dx} (x\, \sin^{-1} y^2) \) \(=0\)  
\(\sin^{-1} y^2+ \dfrac{2xy}{\sqrt{1-y^4}} \cdot \dfrac{dy}{dx} \) \(=0\)  

 
\(\dfrac{dy}{dx} =-\dfrac{\sin^{-1}y^2 \times \sqrt{1-y^4}}{2xy} \)

\(\text{At}\ \ \Big{(} 6, \dfrac{1}{\sqrt{2}} \Big{)}, \)

\(\dfrac{dy}{dx}\) \(= -\dfrac{\sin^{-1}(\frac{1}{2}) \times \sqrt{\frac{3}{4}}}{2 \times 6 \times \frac{1}{\sqrt 2}}\)  
  \(=-\dfrac{\pi}{6} \times \dfrac{\sqrt3}{2} \times \dfrac{\sqrt2}{12} \)  
  \(=-\dfrac{\pi \sqrt6}{144} \)  

Calculus, SPEC1 2023 VCAA 3

A particle moves along a straight line. When the particle is \(x\) m from a fixed point \(O\), its velocity, \( v\) m s\(^{-1}\), is given by

\(v=\dfrac{3 x+2}{2 x-1}\), where  \(x \geq 1\).

  1. Find the acceleration of the particle, in m s\(^{-2}\), when  \(x=2\).   (2 marks)
  1. Find the value that the velocity of the particle approaches as \(x\) becomes very large.   (1 mark)
Show Answers Only

a.    \(a= -\dfrac{56}{27} \)

b.    \(\text{As}\ \ x \rightarrow \infty, \ v \rightarrow \dfrac{3}{2} \)

Show Worked Solution
a.     \(a\) \(=v \cdot \dfrac{dv}{dx}\)
    \(=\dfrac{3x+2}{2x-1} \cdot \Bigg{(} \dfrac{3(2x-1)-2(3x+2)}{(2x-1)^2} \Bigg{)} \)

 
\(\text{When}\ \ x=2: \)

\(a=\dfrac{8}{3} \cdot \dfrac{(9-16)}{9}= -\dfrac{56}{27} \)
 

b.    \(v=\dfrac{3x+2}{2x-1} = \dfrac{3+\frac{2}{x}}{2-\frac{1}{x}} \)

\(\text{As}\ \ x \rightarrow \infty, \ v \rightarrow \dfrac{3}{2} \)

Complex Numbers, SPEC1 2023 VCAA 2

Consider the complex number  \(z=(b-i)^3\), where  \(b \in R^{+}\).

Find \(b\) given that  \(\arg (z)=-\dfrac{\pi}{2}\).   (3 marks)

Show Answers Only

\(b=\sqrt{3} \)

Show Worked Solution

\(-\dfrac{\pi}{2} = \arg (z) = \arg[(b-i)^3]\)

\(b-i\ \Rightarrow \ \text{complex number that lives on}\ \ y=-i \)

\(3 \times \arg(b-i) = -\dfrac{\pi}{2} \)

\(\arg(b-i) = -\dfrac{\pi}{6} \)

\(\tan^{-1}\Big{(}\dfrac{-1}{b} \Big{)} \) \(= -\dfrac{\pi}{6} \)  
\(-\dfrac{1}{b}\) \(=-\dfrac{1}{\sqrt{3}}\)  
\(\therefore b\) \(=\sqrt{3}\)  

Networks, GEN2 2023 VCAA 14

One of the landmarks in state \(A\) requires a renovation project.

This project involves 12 activities, \(A\) to \(L\). The directed network below shows these activities and their completion times, in days.
 

The table below shows the 12 activities that need to be completed for the renovation project.

It also shows the earliest start time (EST), the duration, and the immediate predecessors for the activities.

The immediate predecessor(s) for activity \(I\) and the EST for activity \(J\) are missing.

\begin{array} {|c|c|c|}
\hline
\quad \textbf{Activity} \quad & \quad\quad\textbf{EST} \quad\quad& \quad\textbf{Duration}\quad & \textbf{Immediate} \\
&  & & \textbf{predecessor(s)}  \\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & 0 & 6 & - \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & 0 & 4 & - \\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & 6 & 7 & A \\
\hline
\rule{0pt}{2.5ex} D \rule[-1ex]{0pt}{0pt} & 4 & 5 & B \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & 4 & 10 & B \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & 13 & 4 & C \\
\hline
\rule{0pt}{2.5ex} G \rule[-1ex]{0pt}{0pt} & 9 & 3 & D \\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} & 9 & 7 & D \\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & 13 & 6 & - \\
\hline
\rule{0pt}{2.5ex} J \rule[-1ex]{0pt}{0pt} & - & 6 & E, H \\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} & 19 & 4 & F, I \\
\hline
\rule{0pt}{2.5ex} L \rule[-1ex]{0pt}{0pt} & 23 & 1 & J, K \\
\hline
\end{array}

  1. Write down the immediate predecessor(s) for activity \(I\).  (1 mark)

  2. What is the earliest start time, in days, for activity \(J\) ?  (1 mark)

  3. How many activities have a float time of zero?  (1 mark)

The managers of the project are able to reduce the time, in days, of six activities.

These reductions will result in an increase in the cost of completing the activity.

The maximum decrease in time of any activity is two days.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Activity} \rule[-1ex]{0pt}{0pt} & \quad A \quad & \quad B  \quad&  \quad F \quad  &  \quad H \quad &  \quad I  \quad &  \quad K \quad \\
\hline
\rule{0pt}{2.5ex} \textbf{Daily cost (\$)} \rule[-1ex]{0pt}{0pt} &  1500 & 2000 & 2500 & 1000 & 1500 & 3000  \\
\hline
\end{array}

  1. If activities \(A\) and \(B\) have their completion time reduced by two days each, the overall completion time of the project will be reduced.
  2. What will be the maximum reduction time, in days?  (1 mark)
  3. The managers of the project have a maximum budget of $15 000 to reduce the time for several activities to produce the maximum reduction in the project's overall completion time.
  4. Complete the table below, showing the reductions in individual activity completion times that would achieve the earliest completion time within the $ 15 000 budget.  (1 mark)

\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} &  \\
\hline
\end{array}

Show Answers Only

a.    \(\text{Immediate predecessors of}\ I:\ C, G\)

b.    \(\text{EST}(J) = 4+5+7 = 16\ \text{days}\)

c.    \(\text{5 activities have a float time of zero.}\)

d.    \(\text{Maximum reduction time = 2 days}\)

e.    

\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} &  0\\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} &  1\\
\hline
\end{array}

Show Worked Solution

a.    \(\text{Immediate predecessors of}\ I:\ C, G\)

\(\text{Dummy activity before activity}\ C\ \text{does not effect this.}\)
 

b.    \(\text{Scan network:}\)

\(\text{EST}(J) = 4+5+7 = 16\ \text{days}\)
 

c.    \(\text{Critical Path:}\ A\ C\ I\ K\ L\)

\(\text{Activities on the critical path have a float time of zero.}\)

\(\Rightarrow \ \text{5 activities have a float time of zero.}\)
 

d.    \(\text{If activities}\ A\ \text{and}\ B\ \text{are reduced by 2 days,}\)

\(\text{the critical path remains:}\ A\ C\ I\ K\ L\ \text{(22 days)}\)

\(\text{Maximum reduction time = 2 days}\)

♦♦ Mean mark (c) 35%.
♦♦ Mean mark (d) 33%.

e.    

\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} &  0\\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} &  1\\
\hline
\end{array}

♦♦♦ Mean mark (e) 9%.

Networks, GEN2 2023 VCAA 13

The state \(A\) has nine landmarks, \(G, H, I, J, K, L, M, N\) and \(O\).

The edges on the graph represent the roads between the landmarks.

The numbers on each edge represent the length, in kilometres, along each road.
 

 

Three friends, Eden, Reynold and Shyla, meet at landmark \(G\).

  1. Eden would like to visit landmark \(M\).
  2. What is the minimum distance Eden could travel from \(G\) to \(M\) ?  (1 mark)

  3. Reynold would like to visit all the landmarks and return to \(G\).
  4. Write down a route that Reynold could follow to minimise the total distance travelled.  (1 mark)

  5. Shyla would like to travel along all the roads.
  6. To complete this journey in the minimum distance, she will travel along two roads twice.
  7. Shyla will leave from landmark \(G\) but end at a different landmark.
  8. Complete the following by filling in the boxes provided.  
  9. The two roads that will be travelled along twice are the roads between:  (1 mark)

Show Answers Only

a.    \(\text{Shortest path:}\ G\ K\ I\ M\)

\(\text{Minimum distance}\ = 1.5+1.2+3.2 = 5.9\ \text{km}\)
 

b.   \(\text{Find the shortest Hamiltonian cycle starting at vertex}\ G:\)

\(G\ H\ K\ I\ J\ M\ O\ L\ N\ G\)

\(G\ N\ L\ O\ M\ J\ I\ K\ H\ G\ \text{(reverse of other path)}\)
 

c.   \(\text{vertex}\ L\ \text{and vertex}\ N\)

\(\text{vertex}\ J\ \text{and vertex}\ M\)

Show Worked Solution

a.    \(\text{Shortest path:}\ G\ K\ I\ M\)

\(\text{Minimum distance}\ = 1.5+1.2+3.2 = 5.9\ \text{km}\)
 

b.   \(\text{Find the shortest Hamiltonian cycle starting at vertex}\ G:\)

\(G\ H\ K\ I\ J\ M\ O\ L\ N\ G\)

\(G\ N\ L\ O\ M\ J\ I\ K\ H\ G\ \text{(reverse of other path)}\)

♦ Mean mark (b) 44%.

c.   \(\text{Using an educated guess and check methodology:}\)

\(\text{vertex}\ L\ \text{and vertex}\ N\)

\(\text{vertex}\ J\ \text{and vertex}\ M\)

♦♦♦ Mean mark (c) 12%.

Networks, GEN2 2023 VCAA 12

A country has five states, \(A, B, C, D\) and \(E\).

A graph can be drawn with vertices to represent each of the states.

Edges represent a border shared between two states.

  1. What is the sum of the degrees of the vertices of the graph above?   (1 mark)

  2. Euler's formula, \(v+f=e+2\), holds for this graph.
  3. i. Complete the formula by writing the appropriate numbers in the boxes provided below.   (1 mark)


  4. ii. Complete the sentence by writing the appropriate word in the space provided below.   (1 mark)


    Euler’s formula holds for this graph because the graph is connected and ______________.
  5. The diagram below shows the position of state \(A\) on a map of this country.
  6. The four other states are indicated on the diagram as 1, 2, 3 and 4.
     

  1. Use the information in the graph above to complete the table below. Match the state \((B, C, D\) and \(E)\) with the corresponding state number \((1,2,3\) and 4\()\) given in the map above.   (1 mark)

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \quad \quad \textbf{State} \quad \quad \rule[-1ex]{0pt}{0pt} & \textbf{State Number} \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} D \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} &   \\
\hline
\end{array}
Show Answers Only

a.    \(\text{Sum of degrees}\ = 2+3+4+3+2 = 14\)
 

b.i.   \(\text{Vertices = 5,  Faces = 4,  Edges = 7}\)

\(\Rightarrow 5 + 4 = 7 + 2\)
 

b.ii.     \(\text{Planar}\)
 

c.   

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \quad \quad \textbf{State} \quad \quad \rule[-1ex]{0pt}{0pt} & \textbf{State Number} \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & 3\\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & 2 \\
\hline
\rule{0pt}{2.5ex} D \rule[-1ex]{0pt}{0pt} & 4 \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} &  1 \\
\hline
\end{array}

Show Worked Solution

a.    \(\text{Sum of degrees}\ = 2+3+4+3+2 = 14\)
 

b.i.   \(\text{Vertices = 5,  Faces = 4,  Edges = 7}\)

\(\Rightarrow 5 + 4 = 7 + 2\)
 

b.ii.     \(\text{Planar}\)
 

c.   

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \quad \quad \textbf{State} \quad \quad \rule[-1ex]{0pt}{0pt} & \textbf{State Number} \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & 3\\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & 2 \\
\hline
\rule{0pt}{2.5ex} D \rule[-1ex]{0pt}{0pt} & 4 \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} &  1 \\
\hline
\end{array}

Matrices, GEN2 2023 VCAA 10

Within the circus, there are different types of employees: directors \((D)\), managers \((M)\), performers \((P)\) and sales staff \((S).\) Customers \((C)\) attend the circus.

Communication between the five groups depends on whether they are customers or employees, and on what type of employee they are.

Matrix \(G\) below shows the communication links between the five groups.

\begin{aligned}
&\quad \quad \quad\quad \quad \quad\quad \quad \quad \ \ \textit{receiver}\\
&\quad \quad\quad \quad \quad\quad \quad \quad D \ \ M \ \  P \ \ \  S \ \ \  C \\
& G=\textit{sender} \quad \begin{array}{ccccc}
D\\
M\\
P\\
S\\
C
\end{array}
\begin {bmatrix}
0 & 1 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 & 1 \\
0 & 1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0
\end{bmatrix}\\
&
\end{aligned}

In this matrix:

    • The ' 1 ' in row \(D\), column \(M\) indicates that the directors can communicate directly with the managers.
    • The ' 0 ' in row \(P\), column \(D\) indicates that the performers cannot communicate directly with the directors.
  1. A customer wants to make a complaint to a director.
  2. What is the shortest communication sequence that will successfully get this complaint to a director?  (1 mark)

  3. Matrix \(H\) below shows the number of two-step communication links between each group. Sixteen elements in this matrix are missing.

\begin{aligned}
&\quad \quad \quad\quad \quad \quad\quad \quad \quad \ \ \textit{receiver}\\
&\quad \quad\quad \quad \quad\quad \quad \quad D \quad M \quad P \quad \ S \quad \ C \\
& H=\textit{sender} \quad \begin{array}{ccccc}
D\\
M\\
P\\
S\\
C
\end{array}
\begin {bmatrix} {\displaystyle}
1 & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} \\
0 & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} \\
1 & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} \\
1 & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} \\
0 & 1 & 0 & 0 & 1
\end{bmatrix}\\
&
\end{aligned}

  2.  i. Complete matrix \(H\) above by filling in the missing elements.  (1 mark)
  4. ii. What information do elements \(g_{21}\) and \(h_{21}\) provide about the communication between the circus employees?  (1 mark)

Show Answers Only

a.    \(C\ S\ M\ D\)
 

b.i.  \( \quad \) \(\begin{aligned} & \begin{array}{cccccc}\quad \ \ \ D \ \ M \ \ \ P  \ \ \ S \ \ \ C\end{array} \\ &
\begin{array}{c}D \\ M \\P \\ S \\ C\end{array}
\begin{bmatrix}
1 & 2 & 1 & 2 & 2 \\
0 & 3 & 1 & 2 & 2 \\
1 & 0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 & 1 \\
0 & 1 & 0 & 0 & 1
\end{bmatrix}\\ & \end{aligned}\)

b.ii. \(g_{21}\ \text{indicates managers can communicate directly with directors.}\)

\(h_{21}\ \text{indicates managers, however, cannot communicate with}\)

\(\text{directors through another person who speaks directly to a director.}\)

Show Worked Solution

a.    \(C\ S\ M\ D\)
 

b.i.  \( \quad \) \(\begin{aligned} & \begin{array}{cccccc}\quad \ \ \ D \ \ M \ \ \ P  \ \ \ S \ \ \ C\end{array} \\ &
\begin{array}{c}D \\ M \\P \\ S \\ C\end{array}
\begin{bmatrix}
1 & 2 & 1 & 2 & 2 \\
0 & 3 & 1 & 2 & 2 \\
1 & 0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 & 1 \\
0 & 1 & 0 & 0 & 1
\end{bmatrix}\\ & \end{aligned}\)

b.ii. \(g_{21}\ \text{indicates managers can communicate directly with directors.}\)

\(h_{21}\ \text{indicates managers, however, cannot communicate with}\)

\(\text{directors through another person who speaks directly to a director.}\)

♦♦♦ Mean mark (b)(ii) 25%.

Matrices, GEN2 2023 VCAA 9

The circus is held at five different locations, \(E, F, G, H\) and \(I\).

The table below shows the total revenue for the ticket sales, rounded to the nearest hundred dollars, for the last 20 performances held at each of the five locations.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Location} \rule[-1ex]{0pt}{0pt} & E & F & G & H & I \\
\hline
\rule{0pt}{2.5ex} \textbf{Ticket Sales} \rule[-1ex]{0pt}{0pt} & \$960\ 000 & \$990\ 500 & \$940\ 100 & \$920\ 800 & \$901\ 300 \\
\hline
\end{array}

The ticket sales information is presented in matrix \(R\) below.

\(R=\begin{bmatrix}
960\ 000 & 990\ 500 & 940\ 100 & 920\ 800 & 901\ 300
\end{bmatrix}\)

  1. Complete the matrix equation below that calculates the average ticket sales per performance at each of the five locations.  (1 mark)

\(\begin {bmatrix}\rule{2cm}{0.25mm} \end {bmatrix}\times R = \begin {bmatrix}\rule{2cm}{0.25mm} &\rule{2cm}{0.25mm} &\rule{2cm}{0.25mm} &\rule{2cm}{0.25mm} &\rule{2cm}{0.25mm} \end {bmatrix}\)

The circus would like to increase its total revenue from the ticket sales from all five locations.

The circus will use the following matrix calculation to target the next 20 performances.

\( [t] \times R \times \begin{bmatrix}
1 \\
1 \\
1 \\
1 \\
1
\end{bmatrix}\)

  1. Determine the value of \(t\) if the circus would like to increase its revenue from ticket sales by 25%.  (1 mark)

The circus moves from one location to the next each month. It rotates through each of the five locations, before starting the cycle again.

The following matrix displays the movement between the five locations.

\begin{aligned}
& \quad \ \ \ this \ month\\
& \ \ \ E \ \ \ F \ \ \ G \ \ \ H \ \ \ I \\
& \begin{bmatrix}
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 0
\end{bmatrix} \begin{array}{ll}
E & \\
F\\
G & \ \ next \ month \\
H & \\
I
\end{array}\\
&
\end{aligned}

  1. The circus started in town \(I\).
  2. What is the order in which the circus will visit the five towns?  (1 mark)

Show Answers Only

a.    \(\Big{[} \dfrac{1}{20} \Big{]} \times R = [48\ 000\ \ \ 49\ 525\ \ \ 47\ 005\ \ \ 46\ 040\ \ \ 45\ 065] \)

b.    \(t=1.25\)

c.    \(I\ H\ E\ G\ F\)

Show Worked Solution

a.    \(\Big{[} \dfrac{1}{20} \Big{]} \times R = [48\ 000\ \ \ 49\ 525\ \ \ 47\ 005\ \ \ 46\ 040\ \ \ 45\ 065] \)

 
b.    \(t=1.25\)

 
c.    \(I\ H\ E\ G\ F\)

\(\text{Process: look for a 1 in “this month” column}\ I \)

\(\Rightarrow\ \text{corresponds to “next month” letter}\ H\)

\(\text{Repeat by then looking for a 1 in “this month” column}\ H \)

\(\Rightarrow\ \text{corresponds to “next month” letter}\ E\ \text{etc…}\)

♦♦♦ Mean mark (a) 25%
♦ Mean mark (b) 47%.

Recursion and Finance, GEN2 2023 VCAA 6

Arthur invests $600 000 in an annuity that provides him with a monthly payment of $3973.00.

Interest is calculated monthly.

Three lines of the amortisation table for this annuity are shown below.

\begin{array} {|c|c|}
\hline
\textbf{Payment} & \textbf{Payment} & \textbf{Interest} & \textbf{Principal reduction} & \textbf{Balance} \\
\textbf{number} & \textbf{(\$) } & \textbf{(\$) } & \textbf{(\$) } & \textbf{(\$) }\\
\hline
\rule{0pt}{2.5ex} 0 \rule[-1ex]{0pt}{0pt} & 0.00 & 0.00 & 0.00 & 600\ 000.00 \\
\hline
\rule{0pt}{2.5ex} 1 \rule[-1ex]{0pt}{0pt} & 3973.00 & 2520.00 & 1453.00& 598\ 547.00\\
\hline
\rule{0pt}{2.5ex} 2 \rule[-1ex]{0pt}{0pt} & 3973.00 & 2513.90 & 1459.10 & 597\ 087.90 \\
\hline
\end{array}

  1. The interest rate for the annuity is 0.42% per month.
  2. Determine the interest rate per annum.   (1 mark)

  3. Using the values in the table, complete the next line of the amortisation table.
  4. Write your answers in the spaces provided in the table below.
  5. Round all values to the nearest cent.   (1 mark)

\begin{array} {|c|c|}
\hline
\textbf{Payment} & \textbf{Payment} & \textbf{Interest} & \textbf{Principal reduction} & \textbf{Balance} \\
\textbf{number} & \textbf{(\$) } & \textbf{(\$) } & \textbf{(\$) } & \textbf{(\$) }\\
\hline
\rule{0pt}{2.5ex} 0 \rule[-1ex]{0pt}{0pt} & 0.00 & 0.00 & 0.00 & 600\ 000.00 \\
\hline
\rule{0pt}{2.5ex} 1 \rule[-1ex]{0pt}{0pt} & 3973.00 & 2520.00 & 1453.00& 598\ 547.00\\
\hline
\rule{0pt}{2.5ex} 2 \rule[-1ex]{0pt}{0pt} & 3973.00 & 2513.90 & 1459.10 & 597\ 087.90 \\
\hline
\rule{0pt}{2.5ex} 3 \rule[-1ex]{0pt}{0pt} &  &  &  &  \\
\hline
\end{array}

  1. Let \(V_n\) be the balance of Arthur's annuity, in dollars, after \(n\) months.
  2. Write a recurrence relation in terms of \(V_0, V_{n+1}\) and \(V_n\) that can model the value of the annuity from month to month.   (1 mark)

  3. The amortisation tables above show that the balance of the annuity reduces each month.
  4. If the balance of an annuity remained constant from month to month, what name would be given to this type of annuity?   (1 mark)

Show Answers Only

a.    \(I\% (\text{annual}) = 12 \times 0.42 = 5.04\% \)

b.    \(\text{Row 3 calculations are as follows:}\)

\(\text{Payment}\ = \$3973.00\ \text{(remains constant)}\)

\(\text{Interest}\ = 597\ 087.90 \times 0.0042 = \$2507.77 \)

\(\text{Principal reduction}\ = 3973.00-2507.77 = \$1465.23 \)

\(\text{Balance}\ = 597\ 087.90-1465.23 = \$595\ 622.67\)

c.    \(V_0 = 600\ 000\)

\(V_{n+1} = 1.0042 \times V_n-3973\)

d.    \(\text{Perpetuity}\)

Show Worked Solution

a.    \(I\% (\text{annual}) = 12 \times 0.42 = 5.04\% \)
 

b.    \(\text{Row 3 calculations are as follows:}\)

\(\text{Payment}\ = \$3973.00\ \text{(remains constant)}\)

\(\text{Interest}\ = 597\ 087.90 \times 0.0042 = \$2507.77 \)

\(\text{Principal reduction}\ = 3973.00+2507.77 = \$1465.23 \)

\(\text{Balance}\ = 597\ 087.90-1465.23 = \$595\ 622.67\)

♦♦ Mean mark (b) 40%.

 
c.    \(V_0 = 600\ 000\)

\(V_{n+1} = 1.0042 \times V_n-3973\)

♦ Mean mark (c) 41%.

 
d.    \(\text{Perpetuity}\)

Recursion and Finance, GEN2 2023 VCAA 5

Arthur borrowed $30 000 to buy a new motorcycle.

Interest on this loan is charged at the rate of 6.4% per annum, compounding quarterly.

Arthur will repay the loan in full with quarterly repayments over six years.

  1. How many repayments, in total, will Arthur make?  (1 mark)

The balance of the loan, in dollars, after \(n\) quarters, \(A_n\), can be modelled by the recurrence relation

\(A_0=30\ 000, \quad A_{n+1}=1.016 A_n-1515.18\)

  1. Showing recursive calculations, determine the balance of the loan after two quarters.
  2. Round your answer to the nearest cent.   (1 mark)

  3. The final repayment required will differ slightly from all the earlier repayments of $1515.18
  4. Determine the value of the final repayment.
  5. Round your answer to the nearest cent.   (1 mark)

Show Answers Only

a.    \(\text{Repayments}\ = 6 \times 4 = 24\)

b.    \(A_1=1.016 \times 30\ 000-1515.18=$28\ 964.82 \)

\(A_2=1.016 \times 28\ 964.82-1515.18=$27\ 913.08 \)

c.    \( \text{Final repayment}\ = 1515.18-0.14=\$1515.04\)

Show Worked Solution

a.    \(\text{Repayments}\ = 6 \times 4 = 24\)

  
b.    \(A_1=1.016 \times 30\ 000-1515.18=$28\ 964.82 \)

\(A_2=1.016 \times 28\ 964.82-1515.18=$27\ 913.08 \)

Mean mark (b) 51%.
♦♦ Mean mark (c) 25%

c.    \(\text{Solve for}\ N\ \text{using TMV calculator:}\)

\(N\) \(=24\)  
\(I\%\) \(=6.4\)  
\(PV\) \(= -30\ 000\)  
\(PMT\) \(= 1515.18\)  
\(FV\) \(=?\)  
\(\text{P/Y}\) \(=\ \text{C/Y}\ = 4\)  

 
\(FV = -0.14\)

\(\therefore \text{Final repayment}\ = 1515.18-0.14=\$1515.04\)

Data Analysis, GEN2 2023 VCAA 4

The time series plot below shows the average monthly ice cream consumption recorded over three years, from January 2010 to December 2012.

The data for the graph was recorded in the Northern Hemisphere.

In this graph, month number 1 is January 2010, month number 2 is February 2010 and so on.
 

  1. Identify a feature of this plot that is consistent with this time series having a seasonal component.  (1 mark)

  2. The long-term seasonal index for April is 1.05
  3. Determine the deseasonalised value for average monthly ice cream consumption in April 2010 (month 4).
  4. Round your answer to two decimal places.  (1 mark)

  5. The table below shows the average monthly ice cream consumption for 2011 . 
Consumption (litres/person)
Year Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec
2011 0.156 0.150 0.158 0.180 0.200 0.210 0.183 0.172 0.162 0.145 0.134 0.154
  1. Show that, when rounded to two decimal places, the seasonal index for July 2011 estimated from this data is 1.10.  (2 marks)

Show Answers Only

a.    \(\text{Maximum values are 12 months apart.}\)

b.    \(\text{Deseasonalised (Apr)}\ = \dfrac{0.180}{1.05} = 0.1714… = 0.17\ \text{(2 d.p.)}\)

c.    \(\text{2011 monthly mean}\ =\dfrac{2.004}{12}=0.167\)
 

\(\text{Seasonal index (July)}\) \(=\dfrac{0.183}{0.167}\)  
  \(=1.095…\)  
  \(=1.10\ \text{(2 d.p.)}\)  
Show Worked Solution
a.    \(\text{Maximum values are 12 months apart.}\)
♦♦♦ Mean mark (a) 26%.

 
b.    \(\text{Deseasonalised (Apr)}\ = \dfrac{0.180}{1.05} = 0.1714… = 0.17\ \text{(2 d.p.)}\)

 
c.    \(\text{2011 monthly mean}\)

\(=(0.156+0.15+0.158+0.18+0.2+0.21+0.183+0.172+\)

\(0.162+0.145+0.134+0.154)\ ÷\ 12 \)

\(=\dfrac{2.004}{12}\)

\(=0.167\)
 

\(\text{Seasonal index (July)}\) \(=\dfrac{0.183}{0.167}\)  
  \(=1.095…\)  
  \(=1.10\ \text{(2 d.p.)}\)  
♦♦ Mean mark (c) 36%.

Data Analysis, GEN2 2023 VCAA 3

The scatterplot below plots the average monthly ice cream consumption, in litres/person, against average monthly temperature, in °C. The data for the graph was recorded in the Northern Hemisphere.
 

When a least squares line is fitted to the scatterplot, the equation is found to be:

consumption = 0.1404 + 0.0024 × temperature

The coefficient of determination is 0.7212

  1. Draw the least squares line on the scatterplot graph above.  (1 mark)

  2. Determine the value of the correlation coefficient \(r\).
  3. Round your answer to three decimal places.  (1 mark)

  4. Describe the association between average monthly ice cream consumption and average monthly temperature in terms of strength, direction and form.  (1 mark)

    \begin{array} {|l|c|}
    \hline
    \rule{0pt}{2.5ex} \textbf{strength} \rule[-1ex]{0pt}{0pt} & \quad \quad \quad \quad \quad \quad \quad \quad \\
    \hline
    \rule{0pt}{2.5ex} \textbf{direction} \rule[-1ex]{0pt}{0pt} & \\
    \hline
    \rule{0pt}{2.5ex} \textbf{form} \rule[-1ex]{0pt}{0pt} & \\
    \hline
    \end{array}

  5. Referring to the equation of the least squares line, interpret the value of the intercept in terms of the variables consumption and temperature.  (1 mark)

  6. Use the equation of the least squares line to predict the average monthly ice cream consumption, in litres per person, when the monthly average temperature is –6°C.  (1 mark)

  7. Write down whether this prediction is an interpolation or an extrapolation.  (1 mark)

Show Answers Only

a.    
         

b.    \(r = \sqrt{0.7212} = 0.849\ \text{(3 d.p.)}\)

c.   

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{strength} \rule[-1ex]{0pt}{0pt} & \text{strong} \\
\hline
\rule{0pt}{2.5ex} \textbf{direction} \rule[-1ex]{0pt}{0pt} & \text{positive} \\
\hline
\rule{0pt}{2.5ex} \textbf{form} \rule[-1ex]{0pt}{0pt} & \text{linear} \\
\hline
\end{array}

 
d.    \(\text{At 0°C, the predicted average consumption is:}\)

\(\textit{consumption} = 0.1404 + 0.002 \times 0 = 0.1404\ \text{L/person}\)

 
e.    \(\text{Find consumption} (c)\ \text{when temperature} (t) = -6:\)

\(c=0.1404 + 0.002 \times -6 = 0.1284\ \text{L/person} \)

 
f.     \(\text{Extrapolation (even though the axes extend to –6°C, the data set} \)

\(\text{range finishes with a lower limit around –4.5°C.)}\)

Show Worked Solution

a.    
         

b.    \(r = \sqrt{0.7212} = 0.849\ \text{(3 d.p.)}\)

c.   

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{strength} \rule[-1ex]{0pt}{0pt} & \text{strong} \\
\hline
\rule{0pt}{2.5ex} \textbf{direction} \rule[-1ex]{0pt}{0pt} & \text{positive} \\
\hline
\rule{0pt}{2.5ex} \textbf{form} \rule[-1ex]{0pt}{0pt} & \text{linear} \\
\hline
\end{array}

 
d.    \(\text{At 0°C, the predicted average consumption is:}\)

\(\textit{consumption} = 0.1404 + 0.002 \times 0 = 0.1404\ \text{L/person}\)

 
e.    \(\text{Find consumption} (c)\ \text{when temperature} (t) = -6:\)

\(c=0.1404 + 0.002 \times -6 = 0.1284\ \text{L/person} \)

 
f.     \(\text{Extrapolation (even though the axes extend to –6°C, the data set} \)

\(\text{range finishes with a lower limit around –4.5°C.)}\)

Proof, EXT2 P2 2023 SPEC1 8

A function \(f\) has the rule  \(f(x)=x\,e^{2x}\).

Use mathematical induction to prove that  \(f^{(n)}(x)=\big{(}2^{n}x+n\,2^{n-1}\big{)}e^{2x}\)  for  \(n \in \mathbb{Z}^{+}\), where  \(f^{(n)}(x)\)  represents the \(n\)th derivative of \(f(x)\). That is, \(f(x)\) has been differentiated \(n\) times.   (3 marks)

Show Answers Only

\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\(\text{If}\ \ n=1:\)

\(f(x)=x\,e^{2x}\ \ \Rightarrow \ \ f^{′}(x)=e^{2x} + 2x\,e^{2x} \)

\(f^{(1)}(x)=\big{(}2^{1}x + 1 \times 2^{1-1}\big{)}e^{2x} = e^{2x} + 2x\,e^{2x} \)

\(\therefore\ \text{True for}\ \ n=1.\)
 

\(\text{Assume true for}\ \ n=k \)

\(f^{(k)}(x)=\big{(}2^{k}x + k\,2^{k-1}\big{)}e^{2x}\)
 

\(\text{Prove true for}\ \ n=k+1 \)

\(\text{i.e.}\ f^{(k+1)}(x)=\big{(}2^{k+1}x + (k+1)2^{k}\big{)}e^{2x}\)

\(\text{LHS}\) \(= \dfrac{d}{dx} \big{(}f^{(k)}(x) \big{)} \)   
  \(= \dfrac{d}{dx} \Big{[}\big{(}2^{k}x + k\,2^{k-1}\big{)}e^{2x}\Big{]} \)  
  \(= 2^{k}e^{2x}+(2^{k}x+k\,2^{k-1}) \times 2 e^{2x} \)  
  \(=e^{2x} \big{(}2^{k}+2\cdot2^{k}x + 2k \cdot 2^{k-1}\big{)}\)  
  \(=e^{2x}\big{(}2^{k+1}+2^{k}+k \cdot 2^{k} \big{)} \)  
  \(=\big{(}2^{k+1}x + (k+1)2^{k}\big{)}e^{2x}\)  
  \(=\ \text{RHS}\)  

 
\(\Rightarrow \text{True for}\ \ n=k+1\)

\(\therefore\ \text{Since true for}\ \ n=1\text{, by PMI, true for}\ n \in \mathbb{Z}^{+} \)

Induction, SPEC1 2023 VCAA 8

A function \(f\) has the rule  \(f(x)=x\,e^{2x}\).

Use mathematical induction to prove that  \(f^{(n)}(x)=\big{(}2^{n}x+n\,2^{n-1}\big{)}e^{2x}\)  for  \(n \in \mathbb{Z}^{+}\), where  \(f^{(n)}(x)\)  represents the \(n\)th derivative of \(f(x)\). That is, \(f(x)\) has been differentiated \(n\) times.   (4 marks)

Show Answers Only

\(\text{See Worked Solutions}\)

Show Worked Solution

\(\text{If}\ \ n=1:\)

\(f(x)=x\,e^{2x}\ \ \Rightarrow \ \ f^{′}(x)=e^{2x} + 2x\,e^{2x} \)

\(f^{(1)}(x)=\big{(}2^{1}x + 1 \times 2^{1-1}\big{)}e^{2x} = e^{2x} + 2x\,e^{2x} \)

\(\therefore\ \text{True for}\ \ n=1.\)
 

\(\text{Assume true for}\ \ n=k \)

\(f^{(k)}(x)=\big{(}2^{k}x + k\,2^{k-1}\big{)}e^{2x}\)
 

\(\text{Prove true for}\ \ n=k+1 \)

\(\text{i.e.}\ f^{(k+1)}(x)=\big{(}2^{k+1}x + (k+1)2^{k}\big{)}e^{2x}\)

\(\text{LHS}\) \(= \dfrac{d}{dx} \big{(}f^{(k)}(x) \big{)} \)   
  \(= \dfrac{d}{dx} \Big{[}\big{(}2^{k}x + k\,2^{k-1}\big{)}e^{2x}\Big{]} \)  
  \(= 2^{k}e^{2x}+(2^{k}x+k\,2^{k-1}) \times 2 e^{2x} \)  
  \(=e^{2x} \big{(}2^{k}+2\cdot2^{k}x + 2k \cdot 2^{k-1}\big{)}\)  
  \(=e^{2x}\big{(}2^{k+1}+2^{k}+k \cdot 2^{k} \big{)} \)  
  \(=\big{(}2^{k+1}x + (k+1)2^{k}\big{)}e^{2x}\)  
  \(=\ \text{RHS}\)  

 
\(\Rightarrow \text{True for}\ \ n=k+1\)

\(\therefore\ \text{Since true for n=1, by PMI, true for}\ n \in \mathbb{Z}^{+} \)

Matrices, GEN1 2022 VCAA 5 MC

Matrix \(E\) is a 2 × 2 matrix.

Matrix \(F\) is a 2 × 3 matrix.

Matrix \(G\) is a 3 × 2 matrix.

Matrix \(H\) is a 3 × 3 matrix.

Which one of the following matrix products could have an inverse?

  1. \(EF\)
  2. \(FH\)
  3. \(GE\)
  4. \(GF\)
  5. \(HG\)
Show Answers Only

\(D\)

Show Worked Solution

Matrix product requires: [m × n] × [n × m]

Matrix inverse requires a square matrix: [m × n] × [m × n] = [m × m]

Option A:  \(EF\) = [2 × 2] × [2 × 3] → [2 × 3]  (not square, eliminate A)

Option B:  \(FH\) = [2 × 3] × [3 × 3] → [2 × 3]  (not square, eliminate B)

Option C:  \(GE\) = [3 × 2] × [2 × 2] → [3 × 2]  (not square, eliminate B)

Option D:  \(GF\) = [3 × 2] × [2 × 3] → [3 × 3]  (correct)

Option E:  \(HG\) = [3 × 3] × [3 × 2] → [3 × 2]  (not square, eliminate E)

\(\Rightarrow D\)

Matrices, GEN1 2022 VCAA 3 MC

Each day, members of a swim centre can choose to attend a morning session \((M)\), an afternoon session \((A)\) or no session \((N)\).

The transition diagram below shows the transition from day to day.

The transition diagram is incomplete.
 

Which one of the following transition matrices represents this transition diagram?
 

Show Answers Only

\(C\)

Show Worked Solution

\(\text{By elimination}\)

\(\text{Option E: Eliminate as not all columns add to 1}\)

\(\text{Options A and D: Eliminate as N → M = 0.6, not 0.1 or 0.2}\)

\(\text{Option B: Eliminate as M → A = 0.2, not 0.5}\)

\(\text{Option C: Correct as N → M = 0.6,  M → A = 0.2,  A → N = 0.5 and leading diagonal 0.3, 0.4, 0.1 correct}\)

\(\Rightarrow C\)

Matrices, GEN1 2022 VCAA 1-2 MC

A bike rental business rents road bikes \((R)\) and mountain bikes \((M)\) in three sizes: child \((C)\), junior \((J)\) and adult \((A)\).

Matrix \(B\) shows the daily rental cost, in dollars, for each type of bike.

\begin{aligned} \\
& \quad  R \ \quad \ \  M \\
B = & \begin{bmatrix}
80 & 95  \\
110 & 120 \\
120 & 125
\end{bmatrix}\begin{array}{l}
C\\
J\\
A
\end{array}
\end{aligned}

 
The element in row \(i\) and column \(j\) in matrix \(B\) is \(b_{ij}\).

 
Question 1

The daily cost of renting an adult mountain bike is shown in element

  1. \(b_{12}\)
  2. \(b_{21}\)
  3. \(b_{23}\)
  4. \(b_{31}\)
  5. \(b_{32}\)

 
Question 2

On Sundays, the business increases the daily rental price for each type of bike by 10%.

To determine the rental cost for each type of bike on a Sunday, which one of the following matrix calculations needs to be completed?

  1. \(0.01B\)
  2. \(0.1B\)
  3. \(1.01B\)
  4. \(1.1B\)
  5. \(11B\)
Show Answers Only

\(\text{Question 1:}\ E\)

\(\text{Question 2:}\ D\)

Show Worked Solution

\(\text{Question 1}\)

\(\text{Adult Mountain Bike is in Row 3, Column 2}\ = b_{32}\)

\(\Rightarrow E\)
 

\(\text{Question 2}\)

\(\text{Increase of 10% to each type of bike rental charge.}\)

\(\text{Each element increases by 10%.}\)

\(\text{i.e. Multiply each element by a factor of 1.1.}\)

\(\Rightarrow D\)

Recursion and Finance, GEN1 2022 VCAA 22 MC

Tim deposited $6000 into an investment account earning compound interest calculated monthly.

A rule for the balance, \(T_n\), in dollars, after \(n\) years is given by  \(T_n=6000 \times 1.003^{12n}\).

Let \(R_n\) be a new recurrence relation that models the balance of Tim's account after \(n\) months.

This recurrence relation is

  1. \(R_0=6000,\ \ \ R_{n+1}=R_n+18\)
  2. \(R_0=6000,\ \ \ R_{n+1}=R_n+36\)
  3. \(R_0=6000,\ \ \ R_{n+1}=1.003\,R_n\)
  4. \(R_0=6000,\ \ \ R_{n+1}=1.0036\,R_n\)
  5. \(R_0=6000,\ \ \ R_{n+1}=1.036\,R_n\)
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\(C\)

Show Worked Solution

\(\text{Eliminate option A and B as they are linear.}\)

\(\text{Option D has a monthly interest rate = 0.36% (eliminate)}\)

\(\text{Option E has a monthly interest rate = 3.6% (eliminate)}\)
 

\(\text{Consider option C:}\)

\(T_1 = 6000 \times 1.003^{12 \times 1}\ \ \text{(after 1 year)}\)

\(R_1\) \(= 6000 \times 1.003\) \(= 1.003 \times R_0\)
\(R_2\) \(= 6000 \times 1.003 \times 1.003\) \(= 1.003 \times R_1\)
\(R_3\) \(= 6000 \times 1.003 \times 1.003 \times 1.003\) \(= 1.003 \times R_2\)

  
\(\therefore  R_{n+1} = 1.003\,R_n\ \text{where } n\ \text{is in months.}\)

\(\Rightarrow C\)

Recursion and Finance, GEN1 2022 VCAA 20 MC

Nidhi owns equipment that is used for 10 hours per day for all 365 days of the year.

The value of the equipment is depreciated by Nidhi using the unit cost method.

The value of the equipment, \(E_n\), in dollars, after \(n\) years can be modelled by the recurrence relation

\(E_0=100\ 000,\ \ \ E_{n+1}=E_n-5475\)

The value of the equipment is depreciated by

  1. $1.50 per hour.
  2. $10 per hour.
  3. $15 per hour.
  4. $1.50 per day.
  5. $10 per day.
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\(A\)

Show Worked Solution

\(\text{1st year depreciation = \$5475}\)

\(\text{Hourly depreciation =\dfrac{5475}{365 \times 10}=$1.50\)

\(\Rightarrow A\)

Data Analysis, GEN1 2022 VCAA 15 MC

The daily number of cups of coffee sold by a food truck over a three-week period is shown in the table below.
 

The six-mean smoothed number of cups of coffee, with centring, sold on Thursday in Week 2 is closest to

  1. 127
  2. 138
  3. 147
  4. 155
  5. 163
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\(B\)

Show Worked Solution

\(\text{Six-mean smoothed average (Thu, week 2)}\)

\(=[(98+104+145+163+134+128)\ ÷\ 6 + \)

\((104+145+163+134+128+206)\ ÷\ 6]\ ÷\ 2 \)

\(=137.66…\)

\(\Rightarrow B\)

Data Analysis, GEN1 2022 VCAA 12-14 MC

The scatterplot below displays the body length, in centimetres, of 17 crocodiles, plotted against their head length, in centimetres. A least squares line has been fitted to the scatterplot. The explanatory variable is head length.
 

Question 12

The equation of the least squares line is closest to

  1. head length = –40 + 7 × body length
  2. body length = –40 + 7 × head length
  3. head length = 168 + 7 × body length
  4. body length = 168 – 40 × head length
  5. body length = 7 + 168 × head length

 
Question 13

The median head length of the 17 crocodiles, in centimetres, is closest to

  1. 49
  2. 51
  3. 54
  4. 300
  5. 345

 
Question 14

The correlation coefficient \(r\) is equal to 0.963

The percentage of variation in body length that is not explained by the variation in head length is closest to

  1. 0.9%
  2. 3.7%
  3. 7.3%
  4. 92.7%
  5. 96.3%
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\(\text{Question 12}:\ B\)

\(\text{Question 13}:\ B\)

\(\text{Question 12}:\ C\)

Show Worked Solution

\(\text{Question 12}\)

\(\text{Gradient}\ = \dfrac{550-170}{85-30}=6.9\ \ \text{(eliminate D and E)}\)

\(\text{Head length is the explanatory variable  (eliminate A and C)}\)

\(\Rightarrow  B\)

 
\(\text{Question 13}\)

\(\text{Median score}\ =\dfrac{17+1}{2} = 9\text{th score}\)

\(\text{Median head length = 51 cm}\)

\(\Rightarrow B\)


♦ Mean mark (Q13) 51%.

 
\(\text{Question 14}\)

\(\text{Percentage explained by variation in head length}\)

\(r^2 = 0.963^2 = 0.9273 \approx 92.7\% \)

\(\text{Percentage not explained by variation in head length}\)

\(100-92.7 \approx 7.3\%\) 

\(\Rightarrow C\)


♦ Mean mark (Q14) 48%.

CHEMISTRY, M7 2020 VCE 3

Below is a reaction pathway beginning with hex-3-ene.

  1. Write the IUPAC name of Compound J in the box provided.  (1 mark)

  2. State the reagent(s) required to convert hex-3-ene to hexan-3-ol in the box provided.  (1 mark)

  3. Draw the structural formula for a tertiary alcohol that is an isomer of hexan-3-ol.  (1 mark)

  4. Hexan-3-ol is reacted with Compound M under acidic conditions to produce Compound L.
  5. Draw the semi-structural formula for Compound M in the box provided on the image above.  (1 mark)

  6.  i. Draw the semi-structural formula for Compound K in the box provided on the image above.  (1 mark)

  7. ii. Name the class of organic compound (homologous series) to which Compound K belongs.  (1 mark)

  8. What type of reaction produces Compound K from hexan-3-ol?  (1 mark)

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a.   3-bromohexane

b.   Steam and any specific inorganic strong acid (although not \(\ce{HCl}\)) is correct.

eg. \(\ce{H2O, H+}\)

c.   

 

d.   Correct answers included one of:

\(\ce{CH3COOH}\)  or  \(\ce{HOOCCH3}\)

e.i.  Correct answers included one of the following:

\(\ce{CH3CH2COCH2CH2CH3}\)

\(\ce{CH3CH2CH2COCH2CH3}\)

\(\ce{CH3CH2CO(CH2)2CH3}\)

e.ii.   Ketone

f.    Oxidation

Show Worked Solution

a.   3-bromohexane

b.   Steam and any specific inorganic strong acid (although not \(\ce{HCl}\)) is correct.

eg. \(\ce{H2O, H+}\)
 

♦ Mean mark (b) 38%.

c.   

 

♦ Mean mark (c) 49%.

d.   Correct answers included one of:

\(\ce{CH3COOH}\)  or  \(\ce{HOOCCH3}\)
 

e.i.  Correct answers included one of the following:

\(\ce{CH3CH2COCH2CH2CH3}\)

\(\ce{CH3CH2CH2COCH2CH3}\)

\(\ce{CH3CH2CO(CH2)2CH3}\)
 

e.ii.   Ketone
 

f.    Oxidation

♦ Mean mark e(i) 39%.